Rd Sharma Xi 2020 2021 _volume 2 for Class 11 Science Maths Chapter 29 - Limits

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Rd Sharma Xi 2020 2021 _volume 2 Solutions for Class 11 Science Maths Chapter 29 Limits are provided here with simple step-by-step explanations. These solutions for Limits are extremely popular among Class 11 Science students for Maths Limits Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2020 2021 _volume 2 Book of Class 11 Science Maths Chapter 29 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2020 2021 _volume 2 Solutions. All Rd Sharma Xi 2020 2021 _volume 2 Solutions for class Class 11 Science Maths are prepared by experts and are 100% accurate.

Page No 29.11:

Question 1:

Show that $\lim_{x \to 0} \frac{x}{|x|}$ does not exist.

Answer:

$\lim_{x \to 0} (\frac{x}{|x|})$

Left hand limit:
$\lim_{x \to 0^{-}} (\frac{x}{|x|}) Let x = 0 - h, where h \to 0 . \Rightarrow \lim_{h \to 0} (\frac{0 - h}{|0 - h|}) = \lim_{h \to 0} (\frac{- h}{h}) = - 1$

Right hand limit:
$\lim_{x \to 0^{+}} \frac{(x)}{|x|} Let x = 0 + h, where h \to 0 . \lim_{h \to 0} (\frac{0 + h}{|0 + h|}) = \lim_{h \to 0} (\frac{h}{h}) = 1$

Left hand limit ≠ Right hand limit
$Thus, \lim_{x \to 0} (\frac{x}{|x|}) does not exist .$

Page No 29.11:

Question 2:

Find k so that $\lim_{x \to 2} f (x)$ may exist, where $f (x) = \{\begin{cases} 2 x + 3, & x \leq 2 \\ x + k, & x > 2 \end{cases} .$

Answer:

$f (x) = \{\begin{cases} 2 x + 3, & x \leq 2 \\ x + k, & x > 2 \end{cases} Left hand limit : \lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{-}} (2 x + 3) Let x = 2 - h, where h \to 0 . \lim_{h \to 0} [2 (2 - h) + 3] = 2 (2 - 0) + 3 = 7 Right hand limit : \lim_{x \to 2^{+}} f (x) = \lim_{x \to 2^{+}} (x + k) Let x = 2 + h, where h \to 0 . \lim_{h \to 0} (2 + h + k) = 2 + k$

Now, $\lim_{x \to 2} f (x)$ exists if the left hand limit is equal to the right hand limit.

⇒7 = 2 + k
k = 5

Page No 29.11:

Question 3:

Show that $\lim_{x \to 0} \frac{1}{x}$ does not exist.

Answer:

$\lim_{x \to 0} (\frac{1}{x}) Left hand limit : \lim_{x \to 0^{-}} (\frac{1}{x}) Let x = 0 - h, where h \to 0 \lim_{h \to 0} (\frac{1}{0 - h}) = - \infty Right hand limit : \lim_{x \to 0^{+}} (\frac{1}{x}) Let x = 0 + h, where h \to 0 \lim_{h \to 0} (\frac{1}{0 + h}) = \infty \lim_{x \to 0^{-}} (\frac{1}{x}) \neq \lim_{x \to 0^{+}} (\frac{1}{x}) Thus, \lim_{x \to 0} (\frac{1}{x}) does not exist .$

Page No 29.11:

Question 4:

Let f(x) be a function defined by $f (x) = \{\begin{cases} \frac{3 x}{|x| + 2 x}, & x \neq 0 \\ 0, & x = 0 \end{cases} .$
Show that $\lim_{x \to 0} f (x)$ does not exist.

Answer:

$f (x) = \{\begin{cases} \frac{3 x}{|x| + 2 x}, & x \neq 0 \\ 0, & x = 0 \end{cases} Left hand limit : \lim_{x \to 0^{-}} [\frac{3 x}{|x| + 2 x}] Let x = 0 - h, where h \to 0 . \Rightarrow \lim_{h \to 0} [\frac{3 (- h)}{|- h| + 2 (- h)}] = \lim_{h \to 0} [\frac{- 3 h}{h - 2 h}] = \lim_{h \to 0} [\frac{- 3 h}{- h}] = 3 Right hand limit : \lim_{x \to 0^{+}} (\frac{3 x}{|x| + 2 x}) Let x = 0 + h, where h \to 0 . \Rightarrow^{} \lim_{h \to 0} (\frac{3 h}{|h| + 2 h}) = \lim_{h \to 0} (\frac{3 h}{h + 2 h}) = 1 \lim_{x \to 0^{-}} (\frac{3 x}{|x| + 2 x}) \neq \lim_{x \to 0^{+}} (\frac{3 x}{|x| + 2 x}) Thus, \lim_{x \to 0} f (x) does not exist .$

Page No 29.11:

Question 5:

Let $f (x) = \{\begin{array}{l} x + 1, & if x \geq 0 \\ x - 1, & if x < 0 \end{array} .$ Prove that $\lim_{x \to 0} f (x)$ does not exist.

Answer:

$f (x) = \{\begin{cases} x + 1, & x \geq 0 \\ x - 1, & x < 0 \end{cases} RHL : \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0} (x + 1) Let x = 0 + h, where h \to 0 . \lim_{h \to 0} (0 + h + 1) = 1 LHL : \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (x - 1) Let x = 0 - h, where h \to 0 . \lim_{h \to 0} (0 - h - 1) = - 1 LHL \neq RHL Thus, \lim_{x \to 0} f (x) does not exist .$

Page No 29.11:

Question 6:

Let $f (x) = \{\begin{cases} x + 5, & if x > 0 \\ x - 4, & if x < 0 \end{cases}$ . Prove that $\lim_{x \to 0} f (x)$ does not exist.

Answer:

We have,

$f (x) = \{\begin{cases} x + 5, & if x > 0 \\ x - 4, & if x < 0 \end{cases}$

LHL of f(x) at x = 0

= $\lim_{x \to 0^{-}} f (x) = \lim_{h \to 0} f (0 - h) = \lim_{h \to 0} (0 - h - 4) = - 4$

RHL of f(x) at x = 0

= $\lim_{x \to 0^{+}} f (x) = \lim_{h \to 0} f (0 + h) = \lim_{h \to 0} (0 + h + 5) = 5$

Clearly, $\lim_{x \to 0^{-}} f (x) \neq \lim_{x \to 0^{+}} f (x)$

Hence, $\lim_{x \to 0} f (x)$ does not exist.

Page No 29.11:

Question 7:

Find $\lim_{x \to 3} f (x)$ , where $f (x) = \{\begin{cases} 4, & if x > 3 \\ x + 1, & if x < 3 \end{cases}$

Answer:

We have,

$f (x) = \{\begin{cases} 4, & if x > 3 \\ x + 1, & if x < 3 \end{cases}$

LHL of f(x) at x = 3

= $\lim_{x \to 3^{-}} f (x) = \lim_{h \to 0} f (3 - h) = \lim_{h \to 0} (3 - h + 1) = 4$

RHL of f(x) at x = 3

= $\lim_{x \to 3^{+}} f (x) = \lim_{h \to 0} f (3 + h) = \lim_{h \to 0} 4 = 4$

Clearly, $\lim_{x \to 3^{-}} f (x) = \lim_{x \to 3^{+}} f (x) = 4$

$∴ \lim_{x \to 3} f (x) = 4$

Page No 29.11:

Question 8:

If $f (x) = \{\begin{array}{l} 2 x + 3, & x \leq 0 \\ 3 (x + 1), & x > 0 \end{array} .$ Find $\lim_{x \to 0} f (x)$ and $\lim_{x \to 1} f (x)$ .

Answer:

$f (x) = \{\begin{cases} 2 x + 3, & x \leq 0 \\ 3 (x + 1), & x > 0 \end{cases} LHL : \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (2 x + 3) Let x = 0 - h, where h \to 0 . \lim_{h \to 0} [2 (0 - h) + 3] = \lim_{h \to 0} [- 2 h + 3] = 3 RHL : \lim_{x \to 0^{+}} f (x) \lim_{x \to 0^{+}} 3 (x + 1) Let x = 0 + h, where h \to 0 . \lim_{h \to 0} 3 (0 + h + 1) = \lim_{h \to 0} (3 h + 3) = 3 ∴ LHL = RHL ∴ \lim_{x \to 0} f (x) = 3$

(ii)
$\lim_{x \to 1} f (x) = ? LHL : \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} 3 (x + 1) Let x = 1 - h, where h \to 0 . \lim_{h \to 0} 3 (1 - h + 1) = 6 RHL : \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} 3 (x + 1) Let x = 1 + h, where h \to 0 . \lim_{h \to 0} 3 (1 + h + 1) = 6 ∴ LHL = RHL \lim_{x \to 1} f (x) = 6$

Page No 29.11:

Question 9:

Find $\lim_{x \to 1} f (x)$ , if $f (x) = \{\begin{cases} x^{2} - 1, & x \leq 1 \\ - x^{2} - 1, & x > 1 \end{cases}$

Answer:

$\lim_{x \to 1} f (x) = ? f (x) = \{\begin{cases} x^{2} - 1, & x \leq 1 \\ - x^{2} - 1, & x > 1 \end{cases} LHL : \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (x^{2} - 1) Let x = 1 - h, where h \to 0 . \lim_{h \to 0} ({(1 - h)}^{2} - 1) = 0 RHL : \lim_{x \to 1^{+}} f (x) \lim_{x \to 1^{+}} (- x^{2} - 1) Let x = 1 + h, where h \to 0 . \lim_{h \to 0} (- {(1 + h)}^{2} - 1) = - 2 LHL \neq RHL Thus, \lim_{x \to 1} f (x) d oes not exist .$

Page No 29.11:

Question 10:

Evaluate $\lim_{x \to 0} f (x)$ , where $f (x) = \{\begin{cases} \frac{|x|}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$

Answer:

$f (x) = \{\begin{cases} \frac{|x|}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases} LHL : \lim_{x \to 0^{-}} (\frac{|x|}{x}) Let x = 0 - h, where h \to 0 . \Rightarrow \lim_{h \to 0} (\frac{|0 - h|}{- h}) = \lim_{h \to 0} (\frac{h}{- h}) = - 1 RHL : \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} (\frac{|x|}{x}) Let x = 0 + h, where h \to 0 . \lim_{h \to 0} (\frac{|0 + h|}{0 + h}) = 1 LHL \neq RHL \underset{x \to 0}{Thus, \lim} f (x) does not exist .$

Page No 29.11:

Question 11:

Let a₁, a₂, ..., a_n be fixed real numbers such that
f(x) = (x − a₁) (x − a₂) ... (x − a_n)
What is $\lim_{x \to a_{1}} f (x) ?$ For a ≠ a₁, a₂, ..., a_n. Compute $\lim_{x \to a} f (x) .$

Answer:

$f (x) = (x - a_{1}) (x - a_{2}) . . . (x - a_{n}) \lim_{x \to a_{1}} f (x) = \lim_{x \to a_{1}} [(x - a_{1}) (x - a_{2}) . . . (x - a_{n})] = (a_{1} - a_{1}) (a_{1} - a_{2}) . . . (a_{1} - a_{n}) = 0 \lim_{x \to a} f (x) = \lim_{x \to a} (x - a_{1}) (x - a_{2}) . . . (x - a_{n}) = (a - a_{1}) (a - a_{2}) . . . (a - a_{n})$

Page No 29.11:

Question 12:

Find $\lim_{x \to 1^{+}} (\frac{1}{x - 1}) .$

Answer:

$\lim_{x \to 1^{+}} (\frac{1}{x - 1}) Let x = 1 + h, where h \to 0 . \lim_{h \to 0} (\frac{1}{1 + h - 1}) = \infty$

Page No 29.11:

Question 13:

Evaluate the following one sided limits:

(i)
$\lim_{x \to 2^{+}} \frac{x - 3}{x^{2} - 4}$

(ii)
$\lim_{x \to 2^{-}} \frac{x - 3}{x^{2} - 4}$

(iii)
$\lim_{x \to 0^{+}} \frac{1}{3 x}$

(iv)
$\lim_{x \to - 8^{+}} \frac{2 x}{x + 8}$

(v)
$\lim_{x \to 0^{+}} \frac{2}{x^{1 / 5}}$

(vi)
$\lim_{x \to \frac{π}{2}} \tan x$

(vii)
$\lim_{x \to - π / 2^{+}} \sec x$

(viii)
$\lim_{x \to 0^{-}} \frac{x^{2} - 3 x + 2}{x^{3} - 2 x^{2}}$

(ix)
$\lim_{x \to - 2^{+}} \frac{x^{2} - 1}{2 x + 4}$

(x)
$\lim_{x \to 0^{-}} 2 - \cot x$

(xi)
$\lim_{x \to 0^{-}} 1 + cosec x$

Answer:

(i)
$\lim_{x \to 2^{+}} (\frac{x - 3}{x^{2} - 4}) Let x = 2 + h, where h \to 0 . \Rightarrow \lim_{h \to 0} [\frac{(2 + h) - 3}{{(2 + h)}^{2} - 2^{2}}] = \lim_{h \to 0} [\frac{- 1 + h}{(2 + h - 2) (2 + h + 2)}] = - \infty$

(ii)
$\lim_{x \to 2^{-}} (\frac{x - 3}{x^{2} - 4}) Let x = 2 - h, where h \to 0 . \Rightarrow \lim_{h \to 0} [\frac{2 - h - 3}{{(2 - h)}^{2} - 2^{2}}] = \lim_{h \to 0} [\frac{- h - 1}{(2 - h - 2) (2 - h + 2)}] = \lim_{h \to 0} [\frac{- h - 1}{(- h) (4 - h)}] = \lim_{h \to 0} [\frac{1 + h}{h (4 - h)}] = \infty$

(iii)
$\lim_{x \to 0^{+}} (\frac{1}{3 x}) Let x = 0 + h, where h \to 0 . \Rightarrow \lim_{h \to 0} (\frac{1}{3 h}) = \infty$

(iv)
$\lim_{x \to - 8^{+}} (\frac{2 x}{x + 8}) Let x = - 8 + h, where h \to 0 . \Rightarrow \lim_{h \to 0} (\frac{2 (- 8 + h)}{- 8 + h + 8}) = \lim_{h \to 0} (\frac{- 16 + 2 h}{h}) = - \infty$

(v)
$\lim_{x \to 0^{+}} \frac{2}{x^{\frac{1}{5}}} Let x = 0 + h, where h \to 0 . \Rightarrow \lim_{h \to 0} [\frac{2}{h^{\frac{1}{5}}}] = \infty$

(vi)
$\lim_{x \to {\frac{π}{2}}^{-}} (\tan x) Let x = \frac{π}{2} - h, where h \to 0 . \Rightarrow \lim_{h \to 0} [\tan (\frac{π}{2} - h)] = \lim_{h \to 0} (\cot h) = \infty$

(vii)
$\lim_{x \to - {\frac{π}{2}}^{+}} \sec x Let x = - \frac{π}{2} + h, where h \to 0 . \lim_{h \to 0} \sec (- \frac{π}{2} + h) = \lim_{h \to 0} \sec (\frac{π}{2} - h) [\sec (- θ) = \sec θ] = \lim_{h \to 0} cosec h = \infty$

(viii)
$\lim_{x \to 0^{-}} [\frac{x^{2} - 3 x + 2}{x^{3} - 2 x^{2}}] = \lim_{x \to 0^{-}} [\frac{x^{2} - 2 x - x + 2}{x^{2} (x - 1)}] = \lim_{x \to 0^{-}} [\frac{x (x - 2) - 1 (x - 2)}{x^{2} (x - 2)}] = \lim_{x \to 0^{-}} [\frac{(x - 1) (x - 2)}{x^{2} (x - 2)}] Let x = 0 - h, where h \to 0 . \Rightarrow \lim_{h \to 0} [\frac{(- h - 1)}{{(- h)}^{2}}] = - \infty$

(ix)
$\lim_{x \to - 2^{+}} (\frac{x^{2} - 1}{2 x + 4}) Let x = - 2 + h, where h \to 0 . \lim_{h \to 0} [\frac{{(- 2 + h)}^{2} - 1}{2 (- 2 + h) + 4}] = \frac{4 - 1}{- 4 + 4} = \frac{3}{0} = \infty$

(x)
$\lim_{x \to 0^{-}} (2 - \cot x) Let x = 0 - h, where h \to 0 . \lim_{h \to 0} [2 - \cot (- h)] = \lim_{h \to 0} (2 + \cot h) = 2 + \infty = \infty$

(xi)
$\lim_{x \to 0^{-}} [1 + \cos e c x] Let x = 0 - h, where h \to 0 . \lim_{h \to 0} [1 + cosec (- h)] = \lim_{h \to 0} [1 - cosec h] [cosec (- θ) = - cosec θ] = 1 - \infty = - \infty$

Page No 29.12:

Question 14:

Show that $\lim_{x \to 0} e^{- 1 / x}$ does not exist.

Answer:

$\lim_{x \to 0} e^{- \frac{1}{x}} Left hand limit : \lim_{x \to 0^{-}} (e^{- \frac{1}{x}}) Let x = 0 - h, where h \to 0 . \Rightarrow \lim_{h \to 0} e^{(\frac{- 1}{0 - h})} = \lim_{h \to 0} (e^{\frac{1}{h}}) = \lim_{h \to 0} (e^{\frac{1}{h}}) [When h \to 0, then \frac{1}{h} \to \infty .] = e^{\infty} = \infty Right hand limit : \lim_{x \to 0^{+}} (e^{- \frac{1}{x}}) Let x = 0 + h, where h \to 0 . \lim_{h \to 0} (e^{\frac{- 1}{0 + h}}) = \lim_{h \to 0} (\frac{1}{e^{\frac{1}{h}}}) = \frac{1}{\infty} = 0 \lim_{x \to 0^{-}} e^{- \frac{1}{x}} \neq \lim_{x \to 0^{+}} e^{\frac{1}{x}} Thus, \underset{x \to 0}{limit} e^{- \frac{1}{x}} does not exist .$

Page No 29.12:

Question 15:

Find:

$(i) \lim_{x \to 2} [x] (ii) \lim_{x \to \frac{5}{2}} [x] (iii) \lim_{x \to 1} [x]$

Answer:

(i)

$\lim_{x \to 2} [x] LHL : \lim_{x \to 2^{-}} [x] Let x = 2 - h, where h \to 0 . \lim_{h \to 0} [2 - h] = 1 RHL : \lim_{x \to 2^{+}} [x] Let x = 2 + h, where h \to 0 . \lim_{h \to 0} [2 + h] = 2 LHL \neq RHL Thus, \lim_{x \to 2} [x] does not exist .$

(ii)
$\lim_{x \to \frac{5}{2}} [x] LHL \lim_{x \to {\frac{5}{2}}^{-}} [x] Let x = \frac{5}{2} - h, where h \to 0 . \lim_{h \to 0} [\frac{5}{2} - h] = 2 RHL : \lim_{x \to {\frac{5}{2}}^{+}} [x] Let x = \frac{5}{2} + h, where h \to 0 . \Rightarrow \lim_{h \to 0} [\frac{5}{2} + h] ∴ \lim_{x \to \frac{5}{2}} [x] = 2 = 2 ∴ \lim_{x \to \frac{5}{2}} [x] = 2$

(iii)
$\lim_{x \to 1} [x] LHL : \lim_{x \to 1^{-}} [x] Let x = 1 - h, where h \to 0 . \lim_{h \to 0} [1 - h] = 0 RHL : \lim_{x \to 1^{+}} [x] Let x = 1 + h, where h \to 0 . \lim_{h \to 0} [1 + h] = 1 LHL \neq RHL Thus, \lim_{x \to 1} [x] does not exist .$

Page No 29.12:

Question 16:

Prove that $\lim_{x \to a^{+}} [x] = [a]$ for all a ∈ R. Also, prove that $\lim_{x \to 1^{-}} [x] = 0 .$

Answer:

$\lim_{x \to a^{+}} [x] Let x = a + h, where h \to 0 . \lim_{h \to 0} [a + h] = a \lim_{x \to 1^{-}} [x] Let x = 1 - h, where h \to 0 . \lim_{h \to 0} [1 - h] = 0$

Page No 29.12:

Question 17:

Show that $\lim_{x \to 2^{-}} \frac{x}{[x]} \neq \lim_{x \to 2^{+}} \frac{x}{[x]} .$

Answer:

$\lim_{x \to 2^{-}} \frac{x}{[x]} Let x = 2 - h, where h \to 0 . \lim_{h \to 0} \frac{(2 - h)}{[2 - h]} = \frac{2}{1} \lim_{x \to 2^{+}} \frac{x}{[x]} Let x = 2 + h, where h \to 0 . \lim_{h \to 0} \frac{(2 + h)}{[2 + h]} = \frac{2}{2} = 1 ∴ \lim_{x \to 2^{-}} \frac{x}{[x]} \neq \lim_{x \to 2^{+}} \frac{x}{[x]}$

Page No 29.12:

Question 18:

Find $\lim_{x \to 3^{+}} \frac{x}{[x]} .$ Is it equal to $\lim_{x \to 3^{-}} \frac{x}{[x]} .$

Answer:

$\lim_{x \to 3^{+}} \frac{x}{[x]} Let x = 3 + h, where h \to 0 . \Rightarrow \lim_{h \to 0} (\frac{3 + h}{[3 + h]}) = \frac{3}{3} = 1 Also, \lim_{x \to 3^{-}} (\frac{x}{[x]}) Let x = 3 - h, where h \to 0 . \lim_{h \to 0} (\frac{3 - h}{[3 - h]}) = \frac{3}{2} ∴ \lim_{x \to 3^{-}} \frac{x}{[x]} \neq \lim_{x \to 3^{+}} \frac{x}{[x]}$

Page No 29.12:

Question 19:

Find $\lim_{x \to 5 / 2} [x] .$

Answer:

$\lim_{x \to \frac{5}{2}} [x] LHL : \lim_{x \to {\frac{5}{2}}^{-}} [x] Let x = \frac{5}{2} - h, where h \to 0 . \lim_{h \to 0} [\frac{5}{2} - h] = 2 RHL : \lim_{x \to {\frac{5}{2}}^{+}} [x] Let x = \frac{5}{2} + h, where h \to 0 . \lim_{h \to 0} [\frac{5}{2} + h] = 2$

We know:
$\lim_{x \to \frac{5}{2}} [x] = \lim_{x \to {\frac{5}{2}}^{-}} [x] = \lim_{x \to {\frac{5}{2}}^{+}} [x]$

$∴ \lim_{x \to \frac{5}{2}} [x] = 2$

Page No 29.12:

Question 20:

Evaluate $\lim_{x \to 2} f (x)$ (if it exists), where $f (x) = \{\begin{array}{l} x - [x], & x < 2 \\ 4, & x = 2 \\ 3 x - 5, & x > 2 \end{array} .$

Answer:

$f (x) = \{\begin{matrix} x - [x], & x < 2 \\ 4, & x = 2 \\ 3 x - 5, & x > 2 \end{matrix} LHL : \lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{-}} \{x - [x]\} Let x = 2 - h, where h \to 0 . \Rightarrow \lim_{h \to 0} [(2 - h) - [2 - h]] = 2 - (1) = 1 RHL : \lim_{x \to 2^{+}} f (x) = \lim_{x \to 2^{+}} (3 x - 5) Let x = 2 + h, where h \to 0 . \Rightarrow \lim_{h \to 0} [3 (2 + h) - 5] = 6 - 5 = 1 Here, LHL = RHL = 1 ∴ \lim_{x \to 2} f (x) = 1$

Page No 29.12:

Question 21:

Show that $\lim_{x \to 0} \sin \frac{1}{x}$ does not exist.

Answer:

$\lim_{x \to 0} \sin (\frac{1}{x}) L . H . L \lim_{x \to 0^{-}} f (x) \lim_{x \to 0^{-}} \sin (\frac{1}{x}) Let x = 0 - h where h \to 0 = \lim_{h \to 0} \sin (- \frac{1}{h}) = - \lim_{h \to 0} \sin (\frac{1}{h})$

$\lim_{x \to 0^{-}} f (x)$ = – (An oscillating number that oscillates between –1 and 1)

$Thus, \lim_{x \to 0^{-}} f (x) does not exist . Similarly, \lim_{x \to 0^{+}} f (x) does not exist .$

Page No 29.12:

Question 22:

Let $f (x) = \{\begin{cases} \frac{k \cos x}{π - 2 x}, & where x \neq \frac{π}{2} \\ 3, & where x = \frac{π}{2} \end{cases}$ and if $\lim_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2})$ , find the value of k.

Answer:

We have,

$f (x) = \{\begin{cases} \frac{k \cos x}{π - 2 x}, & where x \neq \frac{π}{2} \\ 3, & where x = \frac{π}{2} \end{cases}$

It is given that,

$\lim_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2}) \Rightarrow \lim_{x \to \frac{π}{2}} \frac{k \cos x}{π - 2 x} = 3 \Rightarrow \frac{k}{2} \times \lim_{x - \frac{π}{2} \to 0} \frac{\sin (\frac{π}{2} - x)}{\frac{π}{2} - x} = 3 \Rightarrow \frac{k}{2} \times \lim_{h \to 0} \frac{\sin (- h)}{- h} = 3 (Put x - \frac{π}{2} = h)$
$\Rightarrow \frac{k}{2} \times \lim_{h \to 0} \frac{\sin h}{h} = 3 [\sin (- θ) = - \sin θ] \Rightarrow \frac{k}{2} = 3 (\lim_{x \to 0} \frac{\sin x}{x} = 1) \Rightarrow k = 6$

Hence, the value of k is 6.

Page No 29.18:

Question 1:

$\lim_{x \to 1} \frac{x^{2} + 1}{x + 1}$

Answer:

$\lim_{x \to 1} (\frac{x^{2} + 1}{x + 1}) = \frac{1^{2} + 1}{1 + 1} = 1$

Page No 29.18:

Question 2:

$\lim_{x \to 0} \frac{2 x^{2} + 3 x + 4}{x^{2} + 3 x + 2}$

Answer:

$\lim_{x \to 0} (\frac{2 x^{2} + 3 x + 4}{x^{2} + 3 x + 2}) = \frac{2 \times 0 + 3 \times 0 + 4}{0 + 3 \times 0 + 2} = \frac{4}{2} = 2$

Page No 29.18:

Question 3:

$\lim_{x \to 3} \frac{\sqrt{2 x + 3}}{x + 3}$

Answer:

$\lim_{x \to 3} [\frac{\sqrt{2 x + 3}}{x + 3}] = \frac{\sqrt{2 \times 3 + 3}}{3 + 3} = \frac{3}{6} = \frac{1}{2}$

Page No 29.18:

Question 4:

$\lim_{x \to 1} \frac{\sqrt{x + 8}}{\sqrt{x}}$

Answer:

$\lim_{x \to 1} (\frac{\sqrt{x + 8}}{\sqrt{x}}) = \frac{\sqrt{1 + 8}}{1} = 3$

Page No 29.18:

Question 5:

$\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}$

Answer:

$\lim_{x \to a} (\frac{\sqrt{x} + \sqrt{a}}{x + a}) = \frac{\sqrt{a} + \sqrt{a}}{a + a} = \frac{2 \sqrt{a}}{2 a} = \frac{1}{\sqrt{a}}$

Page No 29.18:

Question 6:

$\lim_{x \to 1} \frac{1 + {(x - 1)}^{2}}{1 + x^{2}}$

Answer:

$\lim_{x \to 1} [\frac{1 + {(x - 1)}^{2}}{1 + x^{2}}] = \frac{1 + {(1 - 1)}^{2}}{1 + 1^{2}} = \frac{1}{2}$

Page No 29.18:

Question 7:

$\lim_{x \to 0} \frac{x^{2 / 3} - 9}{x - 27}$

Answer:

$\lim_{x \to 0} [\frac{x^{2 / 3} - 9}{x - 27}] = \frac{0 - 9}{0 - 27} = \frac{1}{3}$

Page No 29.18:

Question 8:

$\lim_{x \to 0} 9$

Answer:

$\lim_{x \to 0} (9) = 9$

f(x) = 9 is a constant function.
Its value does not depend on x.

Page No 29.18:

Question 9:

$\lim_{x \to 2} (3 - x)$

Answer:

$\lim_{x \to 2} (3 - x) = 3 - 2 = 1$

Page No 29.18:

Question 10:

$\lim_{x \to - 1} (4 x^{2} + 2)$

Answer:

$\lim_{x \to - 1} (4 x^{2} + 2) = 4 {(- 1)}^{2} + 2 = 4 + 2 = 6$

Page No 29.18:

Question 11:

Evaluate the following limits:

$\lim_{x \to - 1} \frac{x^{3} - 3 x + 1}{x - 1}$

Answer:

$\lim_{x \to - 1} (\frac{x^{3} - 3 x + 1}{x - 1}) = \frac{{(- 1)}^{3} - 3 (- 1) + 1}{- 1 - 1} = \frac{- 1 + 3 + 1}{- 2} = \frac{- 3}{2}$

Page No 29.18:

Question 12:

$\lim_{x \to 0} \frac{3 x + 1}{x + 3}$

Answer:

$\lim_{x \to 0} (\frac{3 x + 1}{x + 3}) = \frac{3 \times 0 + 1}{0 + 3} = \frac{1}{3}$

Page No 29.18:

Question 13:

$\lim_{x \to 3} \frac{x^{2} - 9}{x + 2}$

Answer:

$\lim_{x \to 3} (\frac{x^{2} - 9}{x + 2}) = \frac{3^{2} - 9}{3 + 2} = \frac{9 - 9}{5} = 0$

Page No 29.18:

Question 14:

$\lim_{x \to 0} \frac{a x + b}{c x + d}, d \neq 0$

Answer:

$\lim_{x \to 0} (\frac{a x + b}{c x + d}) = \frac{a \times 0 + b}{c \times 0 + d} = \frac{b}{d}$

Page No 29.23:

Question 1:

$\lim_{x \to - 5} \frac{2 x^{2} + 9 x - 5}{x + 5}$

Answer:

$\lim_{x \to - 5} [\frac{2 x^{2} + 9 x - 5}{x + 5}] It is of the form \frac{0}{0} . \lim_{x \to - 5} [\frac{2 x^{2} + 10 x - x - 5}{x + 5}] = \lim_{x \to - 5} [\frac{2 x (x + 5) - 1 (x + 5)}{x + 5}] = \lim_{x \to - 5} [\frac{(2 x - 1) (x + 5)}{x + 5}] = \lim_{x \to - 5} (2 x - 1) = 2 (- 5) - 1 = - 11$

Page No 29.23:

Question 2:

$\lim_{x \to 3} \frac{x^{2} - 4 x + 3}{x^{2} - 2 x - 3}$

Answer:

$\lim_{x \to 3} [\frac{x^{2} - 4 x + 3}{x^{2} - 2 x - 3}] It is of the form \frac{0}{0} . \lim_{x \to 3} [\frac{x^{2} - x - 3 x + 3}{x^{2} - 3 x + x - 3}] = \lim_{x \to 3} [\frac{(x - 3) (x - 1)}{(x + 1) (x - 3)}] = \lim_{x \to 3} (\frac{x - 1}{x + 1}) = \frac{3 - 1}{3 + 1} = \frac{1}{2}$

Page No 29.23:

Question 3:

$\lim_{x \to 3} \frac{x^{4} - 81}{x^{2} - 9}$

Answer:

$\lim_{x \to 3} [\frac{x^{4} - 81}{x^{2} - 9}] It is of the form \frac{0}{0} . \lim_{x \to 3} [\frac{{(x^{2})}^{2} - 9^{2}}{x^{2} - 9}] = \lim_{x \to 3} [\frac{(x^{2} - 9) (x^{2} + 9)}{x^{2} - 9}] = \lim_{x \to 3} (x^{2} + 9) = 3^{2} + 9 = 18$

Page No 29.23:

Question 4:

$\lim_{x \to 2} \frac{x^{3} - 8}{x^{2} - 4}$

Answer:

$\lim_{x \to 2} [\frac{x^{3} - 8}{x^{2} - 4}] It is of the form \frac{0}{0} . \lim_{x \to 2} [\frac{(x - 2) (x^{2} + 2 x + 4)}{(x - 2) [x + 2]}] [\begin{matrix} ∵ A^{3} - B^{3} = (A - B) (A^{2} + A B + B^{2}) \\ ∵ A^{2} - B^{2} = (A - B) (A + B) \end{matrix}] = \frac{2^{2} + 2 \times 2 + 4}{2 + 2} = \frac{12}{4} = 3$

Page No 29.23:

Question 5:

$\lim_{x \to - 1 / 2} \frac{8 x^{3} + 1}{2 x + 1}$

Answer:

$\lim_{x \to - 1 / 2} [\frac{8 x^{3} + 1}{2 x + 1}] It is of the form \frac{0}{0} . \lim_{x \to - 1 / 2} [\frac{{(2 x)}^{3} + 1}{2 x + 1}] = \lim_{x \to - 1 / 2} [\frac{(2 x + 1) \{{(2 x)}^{2} - 2 x \times 1 + 1^{2}\}}{(2 x + 1)}] [∵ A^{3} + B^{3} = (A + B) (A^{2} - A B + B^{2})] = \lim_{x \to - 1 / 2} [{(2 x)}^{2} - 2 x + 1] = {(2 \times \frac{- 1}{2})}^{2} - 2 \times \frac{- 1}{2} + 1 = 1 + 1 + 1 = 3$

Page No 29.23:

Question 6:

$\lim_{x \to 4} \frac{x^{2} - 7 x + 12}{x^{2} - 3 x - 4}$

Answer:

$\lim_{x \to 4} [\frac{x^{2} - 7 x + 12}{x^{2} - 3 x - 4}] It is of the form \frac{0}{0} . \lim_{x \to 4} [\frac{x^{2} - 3 x - 4 x + 12}{x^{2} - 4 x + x - 4}] = \lim_{x \to 4} [\frac{x (x - 3) - 4 (x - 3)}{x (x - 4) + 1 (x - 4)}] = \lim_{x \to 4} [\frac{(x - 4) (x - 3)}{(x - 4) (x + 1)}] = \frac{4 - 3}{4 + 1} = \frac{1}{5}$

Page No 29.23:

Question 7:

$\lim_{x \to 2} \frac{x^{4} - 16}{x - 2}$

Answer:

$\lim_{x \to 2} [\frac{x^{4} - 16}{x - 2}] It is of the form \frac{0}{0} . \lim_{x \to 2} [\frac{{(x^{2})}^{2} - {(4)}^{2}}{x - 2}] = \lim_{x \to 2} [\frac{(x^{2} - 4) (x^{2} + 4)}{x - 2}] = \lim_{x \to 2} [\frac{(x - 2) (x + 2) (x^{2} + 4)}{x - 2}] = (2 + 2) (2^{2} + 4) = 32$

Page No 29.23:

Question 8:

\lim_{x \to 5} \frac{x^{2} - 9 x + 20}{x^{2} - 6 x + 5}

Answer:

$\lim_{x \to 5} [\frac{x^{2} - 9 x + 20}{x^{2} - 6 x + 5}] It is of the form \frac{0}{0} . \lim_{x \to 5} [\frac{x^{2} - 4 x - 5 x + 20}{x^{2} - x - 5 x + 5}] = \lim_{x \to 5} [\frac{x (x - 4) - 5 (x - 4)}{x (x - 1) - 5 (x - 1)}] = \lim_{x \to 5} [\frac{(x - 5) (x - 4)}{(x - 1) (x - 5)}] = \frac{5 - 4}{5 - 1} = \frac{1}{4}$

Page No 29.23:

Question 9:

$\lim_{x \to - 1} \frac{x^{3} + 1}{x + 1}$

Answer:

$\lim_{x \to - 1} [\frac{x^{3} + 1}{x + 1}] It is of the form \frac{0}{0} . \lim_{x \to - 1} [\frac{x^{3} + 1^{3}}{x + 1}] = \lim_{x \to - 1} [\frac{(x + 1) (x^{2} - x + 1)}{(x + 1)}] = {(- 1)}^{2} - (- 1) + 1 = 1 + 1 + 1 = 3$

Page No 29.23:

Question 10:

$\lim_{x \to 5} \frac{x^{3} - 125}{x^{2} - 7 x + 10}$

Answer:

$\lim_{x \to 5} [\frac{x^{3} - 125}{x^{2} - 7 x + 10}] It is of the form \frac{0}{0} . \lim_{x \to 5} [\frac{x^{3} - 5^{3}}{x^{2} - 2 x - 5 x + 10}] = \lim_{x \to 5} [\frac{(x - 5) (x^{2} + 5 x + 5^{2})}{x (x - 2) - 5 (x - 2)}] [∵ A^{3} - B^{3} = (A - B) (A^{2} + A B + B^{2})] = \lim_{x \to 5} [\frac{(x - 5) (x^{2} + 5 x + 25)}{(x - 2) (x - 5)}] = \frac{5^{2} + 5 \times 5 + 25}{5 - 2} = \frac{75}{3} = 25$

Page No 29.23:

Question 11:

$\lim_{x \to \sqrt{2}} \frac{x^{2} - 2}{x^{2} + \sqrt{2} x - 4}$

Answer:

$\lim_{x \to \sqrt{2}} [\frac{x^{2} - 2}{x^{2} + \sqrt{2} x - 4}] It is of the form \frac{0}{0} . \lim_{x \to \sqrt{2}} [\frac{x^{2} - {(\sqrt{2})}^{2}}{x^{2} + 2 \sqrt{2} x - \sqrt{2} x - 4}] = \lim_{x \to \sqrt{2}} [\frac{(x - \sqrt{2}) (x + \sqrt{2})}{x (x + 2 \sqrt{2}) - \sqrt{2} (x + 2 \sqrt{2})}] = \lim_{x \to \sqrt{2}} [\frac{(x - \sqrt{2}) (x + \sqrt{2})}{(x - \sqrt{2}) (x + 2 \sqrt{2})}] = \frac{\sqrt{2} + \sqrt{2}}{\sqrt{2} + 2 \sqrt{2}} = \frac{2 \sqrt{2}}{3 \sqrt{2}} = \frac{2}{3}$

Page No 29.23:

Question 12:

$\lim_{x \to \sqrt{3}} \frac{x^{2} - 3}{x^{2} + 3 \sqrt{3} x - 12}$

Answer:

$\lim_{x \to \sqrt{3}} [\frac{x^{2} - 3}{x^{2} + 3 \sqrt{3} x - 12}] It is of the form \frac{0}{0} . \lim_{x \to \sqrt{3}} [\frac{x^{2} - {(\sqrt{3})}^{2}}{x^{2} + 4 \sqrt{3} x - \sqrt{3} x - 12}] = \lim_{x \to \sqrt{3}} [\frac{(x - \sqrt{3}) (x + \sqrt{3})}{x (x + 4 \sqrt{3}) - \sqrt{3} (x + 4 \sqrt{3})}] = \lim_{x \to \sqrt{3}} [\frac{(x - \sqrt{3}) (x + \sqrt{3})}{(x - \sqrt{3}) (x + 4 \sqrt{3})}] = \frac{\sqrt{3} + \sqrt{3}}{\sqrt{3} + 4 \sqrt{3}} = \frac{2}{5}$

Page No 29.23:

Question 13:

$\lim_{x \to \sqrt{3}} \frac{x^{4} - 9}{x^{2} + 4 \sqrt{3} x - 15}$

Answer:

$\lim_{x \to \sqrt{3}} [\frac{x^{4} - 9}{x^{2} + 4 \sqrt{3} x - 15}] It is of the form \frac{0}{0} . \lim_{x \to \sqrt{3}} [\frac{{(x^{2})}^{2} - {(3)}^{2}}{x^{2} + 5 \sqrt{3} x - \sqrt{3} x - 15}] = \lim_{x \to \sqrt{3}} [\frac{(x^{2} - 3) (x^{2} + 3)}{x (x + 5 \sqrt{3}) - \sqrt{3} (x + 5 \sqrt{3})}] = \lim_{x \to \sqrt{3}} [\frac{\{x^{2} - {(\sqrt{3})}^{2}\} (x^{2} + 3)}{(x - \sqrt{3}) (x + 5 \sqrt{3})}] = \lim_{x \to \sqrt{3}} [\frac{(x - \sqrt{3}) (x + \sqrt{3}) (x^{2} + 3)}{(x - \sqrt{3}) (x + 5 \sqrt{3})}] = \frac{(\sqrt{3} + \sqrt{3}) (3 + 3)}{\sqrt{3} + 5 \sqrt{3}} = \frac{2 \sqrt{3} \times 6}{6 \sqrt{3}} = 2$

Page No 29.23:

Question 14:

$\lim_{x \to 2} (\frac{x}{x - 2} - \frac{4}{x^{2} - 2 x})$

Answer:

$\lim_{x \to 2} [\frac{x}{x - 2} - \frac{4}{x^{2} - 2 x}] = \lim_{x \to 2} [\frac{x}{x - 2} - \frac{4}{x (x - 2)}] = \lim_{x \to 2} [\frac{x^{2} - 4}{x (x - 2)}] = \lim_{x \to 2} [\frac{(x - 2) (x + 2)}{x (x - 2)}] = \lim_{x \to 2} [\frac{(x + 2)}{x}] = \frac{2 + 2}{2} = 2$

Page No 29.23:

Question 15:

$\lim_{x \to 1} (\frac{1}{x^{2} + x - 2} - \frac{x}{x^{3} - 1})$

Answer:

$\lim_{x \to 1} [\frac{(x^{3} - 1) - x (x^{2} + x - 2)}{(x^{2} + x - 2) (x^{3} - 1)}] = \lim_{x \to 1} [\frac{(x^{3} - 1) - x^{3} - x^{2} + 2 x}{(x^{2} + x - 2) (x^{3} - 1)}] = \lim_{x \to 1} [\frac{- x^{2} + 2 x - 1}{(x^{2} + x - 2) (x - 1) (x^{2} + x + 1)}] = \lim_{x \to 1} [\frac{- (x^{2} - 2 x + 1)}{(x^{2} + x - 2) (x - 1) (x^{2} + x + 1)}] = \lim_{x \to 1} [\frac{- {(x - 1)}^{2}}{(x^{2} + x - 2) (x - 1) (x^{2} + x + 1)}] = \lim_{x \to 1} [\frac{- (x - 1)}{(x^{2} + 2 x - x - 2) (x^{2} + x + 1)}] = \lim_{x \to 1} [\frac{- (x - 1)}{\{x (x + 2) - 1 (x + 2)\} (x^{2} + x + 1)}] = \lim_{x \to 1} [\frac{- (x - 1)}{(x - 1) (x + 2) (x^{2} + x + 1)}] = \frac{- 1}{(1 + 2) (1 + 1 + 1)} = \frac{- 1}{9}$

Page No 29.23:

Question 16:

$\lim_{x \to \sqrt{3}} \frac{x^{4} - 9}{x^{2} + 4 \sqrt{3} x - 15}$

Answer:

$\lim_{x \to 3} [\frac{1}{x - 3} - \frac{2}{x^{2} - 4 x + 3}] = \lim_{x \to 3} [\frac{1}{x - 3} - \frac{2}{x^{2} - 3 x - x + 3}] = \lim_{x \to 3} [\frac{1}{x - 3} - \frac{2}{x (x - 3) - 1 (x - 3)}] = \lim_{x \to 3} [\frac{1}{x - 3} - \frac{2}{(x - 1) (x - 3)}] = \lim_{x \to 3} [\frac{x - 1 - 2}{(x - 3) (x - 1)}] = \lim_{x \to 3} [\frac{1}{x - 1}] = \frac{1}{3 - 1} = \frac{1}{2}$

Page No 29.23:

Question 17:

$\lim_{x \to 2} (\frac{1}{x - 2} - \frac{2}{x^{2} - 2 x})$

Answer:

$\lim_{x \to 2} [\frac{1}{x - 2} - \frac{2}{x^{2} - 2 x}] = \lim_{x \to 2} [\frac{1}{x - 2} - \frac{2}{x (x - 2)}] = \lim_{x \to 2} [\frac{x - 2}{x (x - 2)}] = \lim_{x \to 2} [\frac{1}{x}] = \frac{1}{2}$

Page No 29.23:

Question 18:

$\lim_{x \to 1 / 4} \frac{4 x - 1}{2 \sqrt{x} - 1}$

Answer:

$\lim_{x \to 1 / 4} [\frac{4 x - 1}{2 \sqrt{x} - 1}] It is of the form \frac{0}{0} . \lim_{x \to 1 / 4} [\frac{{(2 \sqrt{x})}^{2} - 1^{2}}{2 \sqrt{x} - 1}] = \lim_{x \to 1 / 4} [\frac{(2 \sqrt{x} - 1) (2 \sqrt{x} + 1)}{(2 \sqrt{x} - 1)}] = 2 \sqrt{\frac{1}{4}} + 1 = 2 \times \frac{1}{2} + 1 = 2$

Page No 29.23:

Question 19:

$\lim_{x \to 4} \frac{x^{2} - 16}{\sqrt{x} - 2}$

Answer:

$\lim_{x \to 4} [\frac{x^{2} - 16}{\sqrt{x} - 2}] It is of the form \frac{0}{0} . \lim_{x \to 4} [\frac{x^{2} - 4^{2}}{\sqrt{x} - 2}] = \lim_{x \to 4} [\frac{(x - 4) (x + 4)}{(\sqrt{x} - 2)}] = \lim_{x \to 4} [\frac{\{{(\sqrt{x})}^{2} - 2^{2}\} (x + 4)}{(\sqrt{x} - 2)}] = \lim_{x \to 4} [\frac{(\sqrt{x} - 2) (\sqrt{x} + 2) (x + 4)}{(\sqrt{x} - 2)}] = (2 + 2) (4 + 4) = 32$

Page No 29.23:

Question 20:

$\lim_{x \to 0} \frac{{(a + x)}^{2} - a^{2}}{x}$

Answer:

$\lim_{x \to 0} [\frac{{(a + x)}^{2} - a^{2}}{x}] It is of the form \frac{0}{0} . \lim_{x \to 0} [\frac{a^{2} + x^{2} + 2 a x - a^{2}}{x}] = \lim_{x \to 0} [\frac{x (x + 2 a)}{x}] = \lim_{x \to 0} [x + 2 a] = 0 + 2 a = 2 a$

Page No 29.23:

Question 21:

$\lim_{x \to 2} (\frac{1}{x - 2} - \frac{4}{x^{3} - 2 x^{2}})$

Answer:

$\lim_{x \to 2} [\frac{1}{x - 2} - \frac{4}{x^{3} - 2 x^{2}}] = \lim_{x \to 2} [\frac{1}{x - 2} - \frac{4}{x^{2} (x - 2)}] = \lim_{x \to 2} [\frac{x^{2} - 4}{x^{2} (x - 2)}] = \lim_{x \to 2} [\frac{(x - 2) (x + 2)}{x^{2} (x - 2)}] = \lim_{x \to 2} [\frac{x + 2}{x^{2}}] = \frac{2 + 2}{2^{2}} = 1$

Page No 29.23:

Question 22:

$\lim_{x \to 3} (\frac{1}{x - 3} - \frac{3}{x^{2} - 3 x})$

Answer:

$\lim_{x \to 3} [\frac{1}{x - 3} - \frac{3}{x^{2} - 3 x}] = \lim_{x \to 3} [\frac{1}{x - 3} - \frac{3}{x (x - 3)}] = \lim_{x \to 3} [\frac{x - 3}{x (x - 3)}] = \lim_{x \to 3} [\frac{1}{x}] = \frac{1}{3}$

Page No 29.23:

Question 23:

$\lim_{x \to 1} (\frac{1}{x - 1} - \frac{2}{x^{2} - 1})$

Answer:

$\lim_{x \to 1} [\frac{1}{x - 1} - \frac{2}{x^{2} - 1}] = \lim_{x \to 1} [\frac{1}{x - 1} - \frac{2}{(x - 1) (x + 1)}] = \lim_{x \to 1} [\frac{x + 1 - 2}{(x - 1) (x + 1)}] = \lim_{x \to 1} [\frac{(x - 1)}{(x - 1) (x + 1)}] = \lim_{x \to 1} [\frac{1}{x + 1}] = \frac{1}{1 + 1} = \frac{1}{2}$

Page No 29.23:

Question 24:

$\lim_{x \to 3} (x^{2} - 9) [\frac{1}{x + 3} + \frac{1}{x - 3}]$

Answer:

$\lim_{x \to 3} [(x^{2} - 9) \{\frac{1}{x + 3} + \frac{1}{x - 3}\}] = \lim_{x \to 3} [(x^{2} - 9) \{\frac{x - 3 + x + 3}{(x + 3) (x - 3)}\}] = \lim_{x \to 3} [(x^{2} - 9) (\frac{2 x}{x^{2} - 9})] = \lim_{x \to 3} (2 x) = 2 \times 3 = 6$

Page No 29.23:

Question 25:

$\lim_{x \to 1} \frac{x^{4} - 3 x^{3} + 2}{x^{3} - 5 x^{2} + 3 x + 1}$

Answer:

p(x) = x⁴ $-$ 3x³ + 2
p(1) = 1 $-$ 3 + 2
= 0
Now, $(x - 1)$ is a factor of p(x).

$x^{3} - 2 x^{2} - 2 x - 2 x - 1 x^{4} - 3 x^{3} + 2 x^{4} - x^{3} - + - 2 x^{3} + 2 - 2 x^{3} + 2 x^{2} + - - 2 x^{2} + 2 - 2 x^{2} + 2 x + - - 2 x + 2 - 2 x + 2 + -___________0$

q(x) = x³ $-$ 5x² + 3x + 1
q(1) = 1 $-$ 5 + 3 + 1
= 0
$Now, (x - 1)$ is a factor of q(x).

$x^{2} - 4 x - 1 x - 1 x^{3} - 5 x^{2} + 3 x + 1 x^{3} - x^{2} - + 4 x^{2} + 3 x + 1 4 x^{2} + 4 x - - - x + 1 - x + 1 + - 0$

$\Rightarrow \lim_{x \to 1} [\frac{x^{4} - 3 x^{3} + 2}{x^{3} - 5 x^{2} + 3 x + 1}] = \lim_{x \to 1} [\frac{(x - 1) (x^{3} - 2 x^{2} - 2 x - 2)}{(x - 1) (x^{2} - 4 x - 1)}] = \frac{(1)^{3} - 2 {(1)}^{2} - 2 (1) - 2}{{(1)}^{2} - 4 \times 1 - 1} = \frac{1 - 2 - 2 - 2}{1 - 4 - 1} = \frac{- 5}{- 4} = \frac{5}{4}$

Page No 29.23:

Question 26:

$\lim_{x \to 2} \frac{x^{3} + 3 x^{2} - 9 x - 2}{x^{3} - x - 6}$

Answer:

$\lim_{x \to 2} [\frac{x^{3} + 3 x^{2} - 9 x - 2}{x^{3} - x - 6}]$

It is of the form $\frac{0}{0} .$

Let p(x) = x³ + 3x² $-$ 9x $-$ 2
p(2) = 8 + 12 $-$ 18 $-$ 2
= 0
Now, $(x - 2)$ is a factor of p(x).

$x^{2} + 5 x + 1 x - 2 x^{3} + 3 x^{2} - 9 x - 2 x^{3} - 2 x^{2} - + 5 x^{2} - 9 x - 2 5 x^{2} - 10 x - + x - 2 x - 2 - + 0$

Let q(x) = x³ – x – 6
q(2) = 8 – 2 – 6
= 0
$Now, (x - 2)$ is a factor of q(x).

$x^{2} + 2 x + 3 x - 2 x^{3} - x - 6 x^{3} - 2 x^{2} - + 2 x^{2} - x - 6 2 x^{2} - 4 x - + 3 x - 6 3 x - 6 - + \times \times \times$

$\Rightarrow \lim_{x \to 2} [\frac{x^{3} + 3 x^{2} - 9 x - 2}{x^{3} - x - 6}] = \lim_{x \to 2} [\frac{(x - 2) (x^{2} + 5 x + 1)}{(x - 2) (x^{2} + 2 x + 3)}] = \lim_{x \to 2} [\frac{x^{2} + 5 x + 1}{x^{2} + 2 x + 3}] = \frac{(2)^{2} + 5 \times 2 + 1}{{(2)}^{2} + 2 \times 2 + 3} = \frac{4 + 10 + 1}{4 + 4 + 3} = \frac{15}{11}$

Page No 29.23:

Question 27:

$\lim_{x \to 1} \frac{1 - x^{- 1 / 3}}{1 - x^{- 2 / 3}}$

Answer:

$\lim_{x \to 1} [\frac{1 - x^{- 1 / 3}}{1 - x^{- 2 / 3}}] It is of the form \frac{0}{0} . \lim_{x \to 1} [\frac{(1 - x^{- 1 / 3})}{{(1)}^{2} - {(x^{- 1 / 3})}^{2}}] = \lim_{x \to 1} [\frac{(1 - x^{- 1 / 3})}{(1 - x^{- 13}) (1 + x^{- 1 / 3})}] = \lim_{x \to 1} [\frac{1}{1 + x^{- 1 / 3}}] = \frac{1}{1 + 1} = \frac{1}{2}$

Page No 29.23:

Question 28:

$\lim_{x \to 3} \frac{x^{2} - x - 6}{x^{3} - 3 x^{2} + x - 3}$

Answer:

$\lim_{x \to 3} [\frac{x^{2} - x - 6}{x^{3} - 3 x^{2} + x - 3}] It is of the form \frac{0}{0} . \lim_{x \to 3} [\frac{x^{2} - 3 x + 2 x - 6}{x^{2} (x - 3) + 1 (x - 3)}] = \lim_{x \to 3} [\frac{x (x - 3) + 2 (x - 3)}{(x^{2} + 1) (x - 3)}] = \lim_{x \to 3} [\frac{(x + 2) (x - 3)}{(x^{2} + 1) (x - 3)}] = \frac{3 + 2}{3^{2} + 1} = \frac{5}{10} = \frac{1}{2}$

Page No 29.23:

Question 29:

$\lim_{x \to - 2} \frac{x^{3} + x^{2} + 4 x + 12}{x^{3} - 3 x + 2}$

Answer:

Let p(x) = x³ + x² + 4x + 12
p(–2) = 0
Thus, x = –2 is the root of p(x).
Now, $(x + 2)$ is a factor of p(x).

$x^{2} - x + 6 x + 2 x^{3} + x^{2} + 4 x + 12 x^{3} + 2 x^{2} - - - x^{2} + 4 x + 12 - x^{2} - 2 x + + 6 x + 12 6 x + 12 - - 0$

p(x) = x³ + x² + 4x + 12
= (x + 2)(x² – x + 6)

Let q(x) = x³ – 3x + 2
q $(- 2)$ = $-$ 8 + 6 + 2
= 0
Thus, x = $-$ 2 is the root of q(x).
Now, $(x + 2)$ is a factor of q(x).

$x^{2} - 2 x + 1 x + 2 x^{3} - 3 x + 2 x^{3} + 2 x^{2} - - - 2 x^{2} - 3 x + 2 - 2 x^{2} - 4 x + + x + 2 x + 2 - - \times \times \times$

q(x) = (x + 2)(x² – 2x + 1)

$\Rightarrow \lim_{x \to - 2} [\frac{x^{3} + x^{2} + 4 x + 12}{x^{3} - 3 x + 2}] = \lim_{x \to - 2} [\frac{(x + 2) (x^{2} - x + 6)}{(x + 2) (x^{2} - 2 x + 1)}] = \frac{(- 2)^{2} - (- 2) + 6}{{(- 2)}^{2} - 2 (- 2) + 1} = \frac{4 + 2 + 6}{4 + 4 + 1} = \frac{12}{9} = \frac{4}{3}$

Page No 29.23:

Question 30:

$\lim_{x \to 1} \frac{x^{3} + 3 x^{2} - 6 x + 2}{x^{3} + 3 x^{2} - 3 x - 1}$

Answer:

Let p(x) = x³ + 3x² $-$ 6x + 2
p(1) = 1 + 3 $-$ 6 + 2
= 0
Now, $(x + 2)$ is a factor of p(x).

$x^{2} + 4 x - 2 x - 1 x^{3} + 3 x^{2} - 6 x + 2 x^{3} - x^{2} - + - 4 x^{2} - 6 x + 2 - 4 x^{2} - 4 x + + - 2 x + 2 - 2 x + 2 + - 0$

p(x) = (x $-$ 1)(x² + 4x $-$ 2)

q(x) = x³ + 3x² $-$ 3x + 2
q(1) = 1 + 3 $-$ 3 $-$ 1
= 0
$Now, (x + 2)$ is a factor of p(x).

$x^{2} + 4 x + 1 x - 1 x^{3} + 3 x^{2} - 3 x - 1 x^{3} - x^{2} - + 4 x^{2} - 3 x - 1 4 x^{2} - 4 x - + x - 1 x - 1 - + 0$

$\Rightarrow \lim_{x \to 1} [\frac{x^{3} + 3 x^{2} - 6 x + 2}{x^{3} + 3 x^{2} - 3 x - 1}] = \lim_{x \to 1} [\frac{(x - 1) (x^{2} + 4 x - 2)}{(x - 1) (x^{2} + 4 x + 1)}] = \frac{(1)^{2} + 4 \times 1 - 2}{{(1)}^{2} + 4 \times 1 + 1} = \frac{1 + 4 - 2}{1 + 4 + 1} = \frac{3}{6} = \frac{1}{2}$

Page No 29.24:

Question 31:

$\lim_{x \to 2} [\frac{1}{x - 2} - \frac{2 (2 x - 3)}{x^{3} - 3 x^{2} + 2 x}]$

Answer:

$\lim_{x \to 2} [\frac{1}{x - 2} - \frac{2 (2 x - 3)}{x^{3} - 3 x^{2} + 2 x}] = \lim_{x \to 2} [\frac{1}{x - 2} - \frac{2 (2 x - 3)}{x (x^{2} - 3 x + 2)}] = \lim_{x \to 2} [\frac{1}{x - 2} - \frac{- 2 (2 x - 3)}{x (x^{2} - 2 x - x + 2)}] = \lim_{x \to 2} [\frac{1}{x - 2} - \frac{2 (2 x - 3)}{x \{x (x - 2) - 1 (x - 2)\}}] = \lim_{x \to 2} [\frac{1}{x - 2} - \frac{2 (2 x - 3)}{x (x - 1) (x - 2)}] = \lim_{x \to 2} [\frac{x (x - 1) - 2 (2 x - 3)}{x (x - 1) (x + 2)}] = \lim_{x \to 2} [\frac{x^{2} - x - 4 x + 6}{x (x - 1) (x + 2)}] = \lim_{x \to 2} [\frac{x^{2} - 5 x + 6}{x (x - 1) (x + 2)}] = \lim_{x \to 2} [\frac{x^{2} - 2 x - 3 x + 6}{x (x - 1) (x - 2)}] = \lim_{x \to 2} [\frac{x (x - 2) - 3 (x - 2)}{x (x - 1) (x - 2)}] = \lim_{x \to 2} [\frac{(x - 3) (x - 2)}{x (x - 1) (x - 2)}] = \frac{2 - 3}{2 (2 - 1)} = - \frac{1}{2}$

Page No 29.24:

Question 32:

$\lim_{x \to 1} \frac{\sqrt{x^{2} - 1} + \sqrt{x - 1}}{\sqrt{x^{2} - 1}}, x > 1$

Answer:

$\lim_{x \to 1} [\frac{\sqrt{x^{2} - 1} + \sqrt{x - 1}}{\sqrt{x^{2} - 1}}] It is of the form \frac{0}{0} . \lim_{x \to 1} [\frac{\sqrt{x^{2} - 1}}{\sqrt{x^{2} - 1}} + \frac{\sqrt{x - 1}}{\sqrt{x^{2} - 1}}] = \lim_{x \to 1} [1 + \frac{\sqrt{x - 1}}{(\sqrt{x - 1}) (\sqrt{x + 1})}] = 1 + \frac{1}{\sqrt{1 + 1}} = 1 + \frac{1}{\sqrt{2}} = \frac{\sqrt{2} + 1}{\sqrt{2}}$

Page No 29.24:

Question 33:

$\lim_{x \to 1} \{\frac{x - 2}{x^{2} - x} - \frac{1}{x^{3} - 3 x^{2} + 2 x}\}$

Answer:

$\lim_{x \to 1} [\frac{x - 2}{x^{2} - x} - \frac{1}{x^{3} - 3 x^{2} + 2 x}] = \lim_{x \to 1} [\frac{x - 2}{x (x - 1)} - \frac{1}{x (x^{2} - 3 x + 2)}] = \lim_{x \to 1} [\frac{x - 2}{x (x - 1)} - \frac{1}{x (x^{2} - 2 x - x + 2)}] = \lim_{x \to 1} [\frac{x - 2}{x (x - 1)} - \frac{1}{x \{x (x - 2) - 1 (x - 2)\}}]$
$= \lim_{x \to 1} [\frac{x - 2}{x (x - 1)} - \frac{1}{x (x - 1) (x - 2)}] = \lim_{x \to 1} [\frac{{(x - 2)}^{2} - 1}{x (x - 1) (x - 2)}]$

$= \lim_{x \to 1} [\frac{(x - 1) (x - 3)}{x (x - 1) (x - 2)}] = \frac{(1 - 3)}{1 (1 - 2)} = \frac{- 2}{- 1} = 2$

Page No 29.24:

Question 34:

Evaluate the following limits:

$\lim_{x \to 1} \frac{x^{7} - 2 x^{5} + 1}{x^{3} - 3 x^{2} + 2}$

Answer:

When x = 1, the expression $\frac{x^{7} - 2 x^{5} + 1}{x^{3} - 3 x^{2} + 2}$ assumes the form $\frac{0}{0}$ . So, (x − 1) is a factor of numerator and denominator.

Using long division method, we get

$x^{7} - 2 x^{5} + 1 = (x - 1) (x^{6} + x^{5} - x^{4} - x^{3} - x^{2} - x - 1)$

and $x^{3} - 3 x^{2} + 2 = (x - 1) (x^{2} - 2 x - 2)$

$∴ \lim_{x \to 1} \frac{x^{7} - 2 x^{5} + 1}{x^{3} - 3 x^{2} + 2} = \lim_{x \to 1} \frac{(x - 1) (x^{6} + x^{5} - x^{4} - x^{3} - x^{2} - x - 1)}{(x - 1) (x^{2} - 2 x - 2)} = \lim_{x \to 1} \frac{x^{6} + x^{5} - x^{4} - x^{3} - x^{2} - x - 1}{x^{2} - 2 x - 2} = \frac{1 + 1 - 1 - 1 - 1 - 1 - 1}{1 - 2 - 2} = \frac{- 3}{- 3} = 1$

Page No 29.28:

Question 1:

$\lim_{x \to 0} \frac{\sqrt{1 + x + x^{2}} - 1}{x}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 + x + x^{2}} - 1}{x}]$

When x = 0, the expression $\frac{\sqrt{1 + x + x^{2}} - 1}{x}$ takes the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{1 + x + x^{2}} - 1) (\sqrt{1 + x + x^{2}} + 1)}{x (\sqrt{1 + x + x^{2}} + 1)}] = \lim_{x \to 0} [\frac{1 + x + x^{2} - 1}{x (\sqrt{1 + x + x^{2}} + 1)}] = \lim_{x \to 0} [\frac{x (1 + x)}{x (\sqrt{1 + x + x^{2}} + 1)}] = \frac{1 + 0}{\sqrt{1 + 0 + 0} + 1} = \frac{1}{2}$

Page No 29.28:

Question 2:

$\lim_{x \to 0} \frac{2 x}{\sqrt{a + x} - \sqrt{a - x}}$

Answer:

$\lim_{x \to 0} [\frac{2 x}{\sqrt{a + x} - \sqrt{a - x}}]$

When x = 0, then the expression $\frac{2 x}{\sqrt{a + x} - \sqrt{a - x}}$ becomes $\frac{0}{0}$ .

Rationalising the denominator:

$\lim_{x \to 0} [\frac{2 x}{(\sqrt{a + x} - \sqrt{a - x})} \times \frac{(\sqrt{a + x} + \sqrt{a - x})}{(\sqrt{a + x} + \sqrt{a - x})}]$

= $\lim_{x \to 0} [\frac{(2 x) (\sqrt{a + x} + \sqrt{a - x})}{(a + x) - (a - x)}]$

= $\lim_{x \to 0} [\frac{2 x (\sqrt{a + x} + \sqrt{a - x})}{2 x}]$

= $\sqrt{a} + \sqrt{a}$
= $2 \sqrt{a}$

Page No 29.28:

Question 3:

$\lim_{x \to 0} \frac{\sqrt{a^{2} + x^{2}} - a}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{a^{2} + x^{2}} - a}{x^{2}}]$

On putting x = 0 in the expression $\sqrt{a^{2} + x^{2}} - a$ , it becomes $\frac{0}{0} .$
Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{a^{2} + x^{2}} - a) (\sqrt{a^{2} + x^{2}} + a)}{x^{2} (\sqrt{a^{2} + x^{2}} + a)}]$

= $\lim_{x \to 0} [\frac{a^{2} + x^{2} - a^{2}}{x^{2} (\sqrt{a^{2} + x^{2}} + a)}]$

= $\frac{1}{\sqrt{a^{2}} + a}$

= $\frac{1}{2 a}$

Page No 29.28:

Question 4:

$\lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{2 x}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 + x} - \sqrt{1 - x}}{2 x}]$

It is of the form $\frac{0}{0} .$

Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{1 + x} - \sqrt{1 - x}) (\sqrt{1 + x} + \sqrt{1 - x})}{2 x (\sqrt{1 + x} + \sqrt{1 - x})}]$

= $\lim_{x \to 0} [\frac{(1 + x) - (1 - x)}{2 x (\sqrt{1 + x} + \sqrt{1 - x})}]$

= $\lim_{x \to 0} [\frac{2 x}{2 x (\sqrt{1 + x} + \sqrt{1 - x})}]$

= $\frac{1}{\sqrt{1 + 0} + \sqrt{1 - 0}}$
= $\frac{1}{2}$

Page No 29.28:

Question 5:

$\lim_{x \to 2} \frac{\sqrt{3 - x} - 1}{2 - x}$

Answer:

$\lim_{x \to 2} [\frac{\sqrt{3 - x} - 1}{2 - x}]$

It is of the form $\frac{0}{0} .$
Rationalising the numerator:

$\lim_{x \to 2} [\frac{(\sqrt{3 - x} - 1) (\sqrt{3 - x} + 1)}{(2 - x) (\sqrt{3 - x} + 1)}]$

$= \lim_{x \to 2} [\frac{3 - x - 1}{(2 - x) (\sqrt{3 - x} + 1)}]$

$= \lim_{x \to 2} [\frac{(2 - x)}{(2 - x) (\sqrt{3 - x} + 1)}]$

$= \frac{1}{\sqrt{3 - 2} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$

Page No 29.28:

Question 6:

$\lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}$

Answer:

$\lim_{x \to 3} [\frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}]$

It is of form $\frac{0}{0} .$

Rationalising the denominator:

$\lim_{x \to 3} [\frac{(x - 3) (\sqrt{x - 2} + \sqrt{4 - x})}{(\sqrt{x - 2} - \sqrt{4 - x}) (\sqrt{x - 2} + \sqrt{4 - x})}]$

= $\lim_{x \to 3} [\frac{(x - 3) (\sqrt{x - 2} + \sqrt{4 - x})}{(x - 2) - (4 - x)}]$

= $\lim_{x \to 3} [\frac{(x - 3) (\sqrt{x - 2} + \sqrt{4 - x})}{2 x - 6}]$

= $\lim_{x \to 3} [\frac{(x - 3) (\sqrt{x - 2} + \sqrt{4 - x})}{2 (x - 3)}]$

= $\frac{\sqrt{3 - 2} + \sqrt{4 - 3}}{2}$

= $\frac{\sqrt{1} + \sqrt{1}}{2}$
= $\frac{2}{2} = 1$

Page No 29.28:

Question 7:

$\lim_{x \to 0} \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}$

Answer:

$\lim_{x \to 0} [\frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}]$

It is of the form $\frac{0}{0}$ .

Rationalising the denominator:

$\lim_{x \to 0} [\frac{x}{(\sqrt{1 + x} - \sqrt{1 - x})} \times \frac{(\sqrt{1 + x} + \sqrt{1 - x})}{(\sqrt{1 + x} + \sqrt{1 - x})}]$

= $\lim_{x \to 0} [\frac{x (\sqrt{1 + x} + \sqrt{1 - x})}{(1 + x) - (1 - x)}]$

= $\lim_{x \to 0} [\frac{x (\sqrt{1 + x} + \sqrt{1 - x})}{2 x}]$

= $\frac{\sqrt{1} + \sqrt{1}}{2}$

= $\frac{2}{2}$
= 1

Page No 29.28:

Question 8:

$\lim_{x \to 1} \frac{\sqrt{5 x - 4} - \sqrt{x}}{x - 1}$

Answer:

$\lim_{x \to 1} [\frac{\sqrt{5 x - 4} - \sqrt{x}}{x - 1}]$

It is of the form $\frac{0}{0} .$
Rationalising the numerator:

$\lim_{x \to 1} [(\frac{\sqrt{5 x - 4} - \sqrt{x}}{x - 1}) (\frac{\sqrt{5 x - 4} + \sqrt{x}}{\sqrt{5 x - 4} + \sqrt{x}})]$

= $\lim_{x \to 1} [\frac{5 x - 4 - x}{(x - 1) (\sqrt{5 x - 4} + \sqrt{x})}]$

= $\lim_{x \to 1} [\frac{4 (x - 1)}{(x - 1) (\sqrt{5 x - 4} + \sqrt{x})}]$

= $\frac{4}{\sqrt{5 - 4} + \sqrt{1}}$

= $\frac{4}{2}$

= 2

Page No 29.28:

Question 9:

$\lim_{x \to 1} \frac{x - 1}{\sqrt{x^{2} + 3 - 2}}$

Answer:

$\lim_{x \to 1} [\frac{x - 1}{\sqrt{x^{2} + 3} - 2}]$

It is of the form $\frac{0}{0}$ .
Rationalising the denominator:

$\lim_{x \to 1} [\frac{(x - 1) (\sqrt{x^{2} + 3} + 2)}{(\sqrt{x^{2} + 3} - 2) (\sqrt{x^{2} + 3} + 2)}]$

= $\lim_{x \to 1} [\frac{(x - 1) (\sqrt{x^{2} + 3} + 2)}{x^{2} + 3 - 4}]$

= $\lim_{x \to 1} [\frac{(x - 1) (\sqrt{x^{2} + 3} + 2)}{(x^{2} - 1)}]$

= $\lim_{x \to 1} [\frac{(x - 1) (\sqrt{x^{2} + 3} + 2)}{(x - 1) (x + 1)}]$

= $\frac{\sqrt{1 + 3} + 2}{1 + 1}$

= $\frac{4}{2}$

= 2

Page No 29.28:

Question 10:

$\lim_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^{2} - 9}$

Answer:

$\lim_{x \to 3} [\frac{\sqrt{x + 3} - \sqrt{16}}{x^{2} - 9}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 3} [\frac{(\sqrt{x + 3} - \sqrt{6}) (\sqrt{x + 3} + \sqrt{6})}{(x^{2} - 9) (\sqrt{x + 3} + \sqrt{6})}]$

= $\lim_{x \to 3} [\frac{(x + 3 - 6)}{(x - 3) (x + 3) (\sqrt{x + 3} + \sqrt{6})}]$

= $\frac{1}{6 \times 2 \sqrt{6}}$

= $\frac{1}{12 \sqrt{6}}$

Page No 29.28:

Question 11:

$\lim_{x \to 1} \frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{2} - 1}$

Answer:

$\lim_{x \to 1} [\frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{2} - 1}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 1} [\frac{(\sqrt{5 x - 4} - \sqrt{x}) (\sqrt{5 x - 4} + \sqrt{x})}{(\sqrt{5 x - 4} + \sqrt{x}) (x^{2} - 1)}]$

= $\lim_{x \to 1} [\frac{5 x - 4 - x}{(\sqrt{5 x - 4} + \sqrt{x}) (x - 1) (x + 1)}]$

= $\lim_{x \to 1} [\frac{4 (x - 1)}{(\sqrt{5 x - 4} + \sqrt{x}) (x - 1) (x + 1)}]$

= $\frac{4}{(\sqrt{5 - 4} + \sqrt{1}) (1 + 1)}$

= $\frac{4}{2 \times 2}$

= 1

Page No 29.28:

Question 12:

$\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 + x} - 1}{x}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{1 + x} - 1) (\sqrt{1 + x} + 1)}{x (\sqrt{1 + x} + 1)}]$

= $\lim_{x \to 0} [\frac{1 + x - 1}{x (\sqrt{1 + x} + 1)}]$

= $\frac{1}{\sqrt{1 + 0} + 1}$

= $\frac{1}{2}$

Page No 29.28:

Question 13:

$\lim_{x \to 2} \frac{\sqrt{x^{2} + 1} - \sqrt{5}}{x - 2}$

Answer:

$\lim_{x \to 2} [\frac{\sqrt{x^{2} + 1} - \sqrt{5}}{x - 2}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 2} [\frac{(\sqrt{x^{2} + 1} - \sqrt{5}) (\sqrt{x^{2} + 1} + \sqrt{5})}{(x - 2) (\sqrt{x^{2} + 1} + \sqrt{5})}]$

= $\lim_{x \to 2} [\frac{x^{2} + 1 - 5}{(x - 2) (\sqrt{x^{2} + 1} + \sqrt{5})}]$

= $\lim_{x \to 2} [\frac{x^{2} - 4}{(x - 2) (\sqrt{x^{2} + 1} + \sqrt{5})}]$

= $\lim_{x \to 2} [\frac{(x - 2) (x + 2)}{(x - 2) (\sqrt{x^{2} + 1} + \sqrt{5})}]$

= $\frac{4}{2 \sqrt{5}}$

= $\frac{2}{\sqrt{5}}$

Page No 29.28:

Question 14:

$\lim_{x \to 2} \frac{x - 2}{\sqrt{x} - \sqrt{2}}$

Answer:

$\lim_{x \to 2} [\frac{x - 2}{\sqrt{x} - \sqrt{2}}]$

It is of the form $\frac{0}{0}$ .

⇒ $\lim_{x \to 2} [\frac{{(\sqrt{x})}^{2} - {(\sqrt{2})}^{2}}{x - \sqrt{2}}]$

= $\lim_{x \to 2} [\frac{(\sqrt{x} - \sqrt{2}) (\sqrt{x} + \sqrt{2})}{(x - \sqrt{2})}]$

= $\sqrt{2} + \sqrt{2}$

= $2 \sqrt{2}$

Page No 29.28:

Question 15:

$\lim_{x \to 7} \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}$

Answer:

$\lim_{x \to 7} [\frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator and the denominator:

$\lim_{x \to 7} [\frac{(4 - \sqrt{9 + x})}{1} \times \frac{(4 + \sqrt{9 + x})}{(4 + \sqrt{9 + x})} \times \frac{1}{(1 - \sqrt{8 - x})} \times \frac{(1 + \sqrt{8 - x})}{(1 + \sqrt{8 - x})}]$

= $\lim_{x \to 7} [\frac{16 - (9 + x)}{(4 + \sqrt{9 + x})} \times \frac{(1 + \sqrt{8 - x})}{1 - (8 - x)}]$

= $\lim_{x \to 7} [\frac{- 1 (- 7 + x) (1 + \sqrt{8 - x})}{(4 + \sqrt{9 + x}) (- 7 + x)}]$

= $\lim_{x \to 7} [\frac{- (1 + \sqrt{8 - x})}{4 + \sqrt{9 + x}}]$

= $- (\frac{1 + \sqrt{8 - 7}}{4 + \sqrt{9 + 7}})$

= $\frac{- 2}{4 + 4}$

= $\frac{- 1}{4}$

Page No 29.28:

Question 16:

$\lim_{x \to 0} \frac{\sqrt{a + x} - \sqrt{a}}{x \sqrt{a^{2} + a x}}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{a + x} - \sqrt{a}}{x \sqrt{a^{2} + a x}}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{a + x} - \sqrt{a}) \times (\sqrt{a + x} + \sqrt{a})}{(x \sqrt{a^{2} + a x}) \times (\sqrt{a + x} + \sqrt{a})}]$

= $\lim_{x \to 0} [\frac{a + x - a}{x (\sqrt{a^{2} + a x}) (\sqrt{a + x} + \sqrt{a})}]$

= $\frac{1}{(\sqrt{a^{2}}) (\sqrt{a} + \sqrt{a})}$

= $\frac{1}{2 a \sqrt{a}}$

Page No 29.28:

Question 17:

$\lim_{x \to 5} \frac{x - 5}{\sqrt{6 x - 5} - \sqrt{4 x + 5}}$

Answer:

$\lim_{x \to 5} [\frac{x - 5}{\sqrt{6 x - 5} - \sqrt{4 x + 5}}]$

It is of the form $\frac{0}{0}$ .
Rationalising the denominator:

$\lim_{x \to 5} [\frac{(x - 5) (\sqrt{6 x - 5} + \sqrt{4 x + 5})}{(\sqrt{6 x - 5} - \sqrt{4 x + 5}) (\sqrt{6 x - 5} + \sqrt{4 x + 5})}]$

= $\lim_{x \to 5} [\frac{(x - 5) (\sqrt{6 x - 5} + \sqrt{4 x + 5})}{(6 x - 5) - (4 x + 5)}]$

= $\lim_{x \to 5} [\frac{(x - 5) (\sqrt{6 x - 5} + \sqrt{4 x + 5})}{2 x - 10}]$

= $\lim_{x \to 5} [\frac{(x - 5) (\sqrt{6 x - 5} + \sqrt{4 x + 5})}{2 (x - 5)}]$

= $\frac{\sqrt{6 \times 5 - 5} + \sqrt{4 \times 5 + 5}}{2}$

= $\frac{5 + 5}{2}$

= 5

Page No 29.28:

Question 18:

$\lim_{x \to 1} \frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{3} - 1}$

Answer:

$\lim_{x \to 1} [\frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{3} - 1}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 1} [\frac{(\sqrt{5 x - 4} - \sqrt{x}) (\sqrt{5 x - 4} + \sqrt{x})}{(x^{3} - 1) (\sqrt{5 x - 4} + \sqrt{x})}]$

= $\lim_{x \to 1} [\frac{5 x - 4 - x}{(x - 1) (x^{2} + x + 1) (\sqrt{5 x - 4} + \sqrt{x})}]$

= $\lim_{x \to 1} [\frac{4 (x - 1)}{(x - 1) (x^{2} + x + 1) (\sqrt{5 x - 4} + \sqrt{x})}]$

= $\frac{4}{3 (1 + 1)}$

= $\frac{2}{3}$

Page No 29.28:

Question 19:

$\lim_{x \to 2} \frac{\sqrt{1 + 4 x} - \sqrt{5 + 2 x}}{x - 2}$

Answer:

$\lim_{x \to 2} [\frac{\sqrt{1 + 4 x} - \sqrt{5 + 2 x}}{x - 2}]$

It is of the from $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 2} [\frac{(\sqrt{1 + 4 x} - \sqrt{5 + 2 x}) (\sqrt{1 + 4 x} + \sqrt{5 + 2 x})}{(x - 2) (\sqrt{1 + 4 x} + \sqrt{5 + 2 x})}]$

= $\lim_{x \to 2} [\frac{(1 + 4 x) - (5 + 2 x)}{(x - 2) (\sqrt{1 + 4 x} + \sqrt{5 + 2 x})}]$

= $\lim_{x \to 2} [\frac{2 (x - 2)}{(x - 2) (\sqrt{1 + 4 x} + \sqrt{5 + 2 x})}]$

= $\frac{2}{(\sqrt{1 + 4 \times 2} + \sqrt{5 + 2 \times 2})}$

= $\frac{2}{3 + 3}$

= $\frac{1}{3}$

Page No 29.28:

Question 20:

$\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^{2} - 1}$

Answer:

$\lim_{x \to 1} [\frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^{2} - 1}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 1} [\frac{(\sqrt{3 + x} - \sqrt{5 - x}) (\sqrt{3 + x} + \sqrt{5 - x})}{(x - 1) (x + 1) (\sqrt{3 + x} + \sqrt{5 - x})}]$

= $\lim_{x \to 1} [\frac{(3 + x) - (5 - x)}{(x - 1) (x + 1) \{\sqrt{3 + x} + \sqrt{5 - x}\}}]$

= $\lim_{x \to 1} [\frac{2 x - 2}{(x - 1) (x + 1) \{\sqrt{3 + x} + \sqrt{5 - x}\}}]$

= $\lim_{x \to 1} [\frac{2 (x - 1)}{(x - 1) (x + 1) \{\sqrt{3 + x} + \sqrt{5 - x}\}}]$

= $\frac{2}{2 (\sqrt{4} + \sqrt{4})}$

= $\frac{1}{4}$

Page No 29.29:

Question 21:

$\lim_{x \to 0} \frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{x}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{x}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}) (\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}})}{(\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}) x}]$

= $\lim_{x \to 0} [\frac{(1 + x^{2}) - (1 - x^{2})}{x \{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}\}}]$

= $\lim_{x \to 0} [\frac{2 x^{2}}{x \{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}\}}]$

= $\frac{2 \times 0}{\sqrt{1 + 0} + \sqrt{1 - 0}}$

= 0

Page No 29.29:

Question 22:

$\lim_{x \to 0} \frac{\sqrt{1 + x + x^{2}} - \sqrt{x + 1}}{2 x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 + x + x^{2}} - \sqrt{x + 1}}{2 x^{2}}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{1 + x + x^{2}} - \sqrt{x + 1}) (\sqrt{1 + x + x^{2}} + \sqrt{x + 1})}{(\sqrt{1 + x + x^{2}} + \sqrt{x + 1}) 2 x^{2}}]$

= $\lim_{x \to 0} [\frac{(1 + x + x^{2}) - (x + 1)}{(\sqrt{1 + x + x^{2}} + \sqrt{x + 1}) 2 x^{2}}]$

= $\lim_{x \to 0} [\frac{x^{2}}{(\sqrt{1 + x + x^{2}} + \sqrt{x + 1}) (2 x^{2})}]$

= $\frac{1}{(\sqrt{1 + 0 + 0} + \sqrt{0 + 1}) \times 2}$

= $\frac{1}{2} \times \frac{1}{2}$

= $\frac{1}{4}$

Page No 29.29:

Question 23:

$\lim_{x \to 4} \frac{2 - \sqrt{x}}{4 - x}$

Answer:

$\lim_{x \to 4} [\frac{2 - \sqrt{x}}{4 - x}]$

= $\lim_{x \to 4} [\frac{2 - \sqrt{x}}{2 - {(\sqrt{x})}^{2}}]$

= $\lim_{x \to 4} [\frac{(2 - \sqrt{x})}{(2 - \sqrt{x}) (2 + \sqrt{x})}]$

= $\frac{1}{2 + \sqrt{4}}$

= $\frac{1}{2 + 2}$

= $\frac{1}{4}$

Page No 29.29:

Question 24:

$\lim_{x \to a} \frac{x - a}{\sqrt{x} - \sqrt{a}}$

Answer:

$\lim_{x \to a} [\frac{x - a}{\sqrt{x} - \sqrt{a}}]$

= $\lim_{x \to a} [\frac{{(\sqrt{x})}^{2} - a^{2}}{\sqrt{x} - \sqrt{a}}]$

= $\lim_{x \to a} [\frac{(\sqrt{x} - \sqrt{a}) (\sqrt{x} + \sqrt{a})}{(\sqrt{x} - \sqrt{a})}]$

= $\sqrt{a} + \sqrt{a}$

= $2 \sqrt{a}$

Page No 29.29:

Question 25:

$\lim_{x \to 0} \frac{\sqrt{1 + 3 x} - \sqrt{1 - 3 x}}{x}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 + 3 x} - \sqrt{1 - 3 x}}{x}]$

It is of the form $\frac{0}{0}$ .

Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{1 + 3 x} - \sqrt{1 - 3 x}) (\sqrt{1 + 3 x} + \sqrt{1 - 3 x})}{x (\sqrt{1 + 3 x} + \sqrt{1 - 3 x})}] = \lim_{x \to 0} [\frac{(1 + 3 x) - (1 - 3 x)}{x (\sqrt{1 + 3 x} + \sqrt{1 - 3 x})}] = \lim_{x \to 0} [\frac{6 x}{x (\sqrt{1 + 3 x} + \sqrt{1 - 3 x})}] = \frac{6}{\sqrt{1} + \sqrt{1}} = \frac{6}{2} = 3$

Page No 29.29:

Question 26:

$\lim_{x \to 0} \frac{\sqrt{2 - x} - \sqrt{2 + x}}{x}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{2 - x} - \sqrt{2 + x}}{x}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{2 - x} - \sqrt{2 + x}) (\sqrt{2 - x} + \sqrt{2 + x})}{x (\sqrt{2 - x} + \sqrt{2 + x})}]$

= $\lim_{x \to 0} [\frac{(2 - x) - (2 + x)}{x (\sqrt{2 - x} + \sqrt{2 + x})}]$

= $\lim_{x \to 0} [\frac{- 2 x}{x (\sqrt{2 - x} + \sqrt{2 + x})}]$

= $\frac{- 2}{2 \sqrt{2}}$

= $- \frac{1}{\sqrt{2}}$

Page No 29.29:

Question 27:

$\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^{2} - 1}$

Answer:

$\lim_{x \to 1} [\frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^{2} - 1}]$

It is of the form $\frac{0}{0}$ .

Rationalising the numerator:

$\lim_{x \to 1} [\frac{(\sqrt{3 + x} - \sqrt{5 - x}) (\sqrt{3 + x} + \sqrt{5 - x})}{(x - 1) (x + 1) (\sqrt{3 + x} + \sqrt{5 - x})}] = \lim_{x \to 1} [\frac{(3 + x) - (5 - x)}{(x - 1) (x + 1) (\sqrt{3 + x} + \sqrt{5 - x})}] = \lim_{x \to 1} \frac{2 (x - 1)}{(x - 1) (x + 1) (\sqrt{3 + x} + \sqrt{5 - x})} = \frac{2}{(1 + 1) (\sqrt{3 + 1} + \sqrt{5 - 1})} = \frac{2}{2 \times (2 + 2)} = \frac{1}{4}$

Page No 29.29:

Question 28:

$\lim_{x \to 1} \frac{(2 x - 3) (\sqrt{x} - 1)}{3 x^{2} + 3 x - 6}$

Answer:

$\lim_{x \to 1} [\frac{(2 x - 3) (\sqrt{x} - 1)}{3 x^{2} + 3 x - 6}]$

It is of the form $\frac{0}{0}$ .

⇒ $\lim_{x \to 1} [\frac{(2 x - 3) (\sqrt{x} - 1)}{3 (x^{2} + x - 2)}]$

= $\lim_{x \to 1} [\frac{(2 x - 3) (\sqrt{x} - 1)}{3 (x^{2} + 2 x - x - 2)}]$

= $\lim_{x \to 1} [\frac{(2 x - 3) (\sqrt{x} - 1)}{3 (x (x + 2) - 1 (x + 2))}]$

= $\lim_{x \to 1} [\frac{(2 x - 3) (\sqrt{x} - 1)}{3 (x - 1) (x + 2)}]$

= $\lim_{x \to 1} [\frac{(2 x - 3) (\sqrt{x} - 1)}{3 ({(\sqrt{x})}^{2} - 1^{2}) (x + 2)}]$

= $\lim_{x \to 1} [\frac{(2 x - 3) (\sqrt{x} - 1)}{3 (\sqrt{x} + 1) (\sqrt{x} - 1) (x + 2)}]$

= $\frac{- 1}{3 (2) \times 3}$

= $\frac{- 1}{18}$

Page No 29.29:

Question 29:

$\lim_{x \to 0} \frac{\sqrt{1 + x^{2}} - \sqrt{1 + x}}{\sqrt{1 + x^{3}} - \sqrt{1 + x}}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 - x^{2}} - \sqrt{1 + x}}{\sqrt{1 + x^{3}} - \sqrt{1 + x}}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator and the denominator:

$\lim_{x \to 0} [\frac{(\sqrt{1 + x^{2}} - \sqrt{1 + x})}{1} \times \frac{(\sqrt{1 + x^{2}} + \sqrt{1 + x})}{(\sqrt{1 + x^{2}} + \sqrt{1 + x})} \times \frac{1}{(\sqrt{1 + x^{3}} - \sqrt{1 + x})} \times \frac{(\sqrt{1 + x^{3}} - \sqrt{1 + x})}{(\sqrt{1 + x^{3}} + \sqrt{1 + x})}] = \lim_{x \to 0} [\frac{[(1 + x^{2}) - (1 + x)]}{[(1 + x^{3}) - (1 + x)]} \times \frac{(\sqrt{1 + x^{3}} + \sqrt{1 + x})}{(\sqrt{1 + x^{2}} + \sqrt{1 + x})}] = \lim_{x \to 0} [(\frac{x^{2} - x}{x^{3} - x}) \times \frac{(\sqrt{1 + x^{3}} + \sqrt{1 + x})}{(\sqrt{1 + x^{2}} + \sqrt{1 + x})}] = \lim_{x \to 0} [\frac{x (x - 1)}{x (x^{2} - 1)} \frac{(\sqrt{1 + x^{3}} + \sqrt{1 + x})}{(\sqrt{1 + x^{2}} + \sqrt{1 + x})}] = \lim_{x \to 0} [\frac{(x - 1) (\sqrt{1 + x^{3}} + \sqrt{1 + x})}{(x - 1) (x + 1) (\sqrt{1 + x^{2}} + \sqrt{1 + x})}] = \frac{(\sqrt{1 + 0} + \sqrt{1 + 0})}{(0 + 1) (\sqrt{1 + 0} + \sqrt{1 + 0})} = \frac{2}{2} = 1$

Page No 29.29:

Question 30:

$\lim_{x \to 1} \frac{x^{2} - \sqrt{x}}{\sqrt{x} - 1}$

Answer:

$\lim_{x \to 1} [\frac{x^{2} - \sqrt{x}}{\sqrt{x} - 1}]$

It is of the form $\frac{0}{0}$ .

$\lim_{x \to 1} [\frac{\sqrt{x} (x \sqrt{x} - 1)}{\sqrt{x} - 1}]$

= $\lim_{x \to 1} [\frac{\sqrt{x} (x^{3 / 2} - 1)}{x^{\frac{1}{2}} - 1}]$

= $\lim_{x \to 1} [\frac{\sqrt{x} ({(\sqrt{x})}^{3} - 1^{3})}{(\sqrt{x} - 1)}] [A^{3} - B^{3} = (A - B) (A^{2} + A B + B^{2})]$

= $\lim_{x \to 1} [\frac{(\sqrt{x}) (\sqrt{x} - 1) (x + \sqrt{x} + 1)}{(\sqrt{x} - 1)}]$

= 1 (1 + 1 + 1)

= 3

Page No 29.29:

Question 31:

$\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}, x \neq 0$

Answer:

$\lim_{h \to 0} [\frac{\sqrt{x + h} - \sqrt{x}}{h}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:
= $\lim_{h \to 0} [\frac{(\sqrt{x + h} - \sqrt{x})}{h} \times \frac{(\sqrt{x + h} + \sqrt{x})}{(\sqrt{x + h} + \sqrt{x})}]$

= $\lim_{h \to 0} [\frac{(x + h) - x}{h (\sqrt{x + h} + \sqrt{x})}]$

= $\frac{1}{\sqrt{x} + \sqrt{x}}$

= $\frac{1}{2 \sqrt{x}}$

Page No 29.29:

Question 32:

$\lim_{x \to \sqrt{10}} \frac{\sqrt{7 + 2 x} - (\sqrt{5} + \sqrt{2})}{x^{2} - 10}$

Answer:

$\lim_{x \to \sqrt{10}} [\frac{\sqrt{7 + 2 x} - (\sqrt{5} + \sqrt{2})}{x^{2} - {(\sqrt{10})}^{2}}]$

= $\lim_{x \to \sqrt{10}} [\frac{\sqrt{7 + 2 x} - \sqrt{{(\sqrt{5} + \sqrt{2})}^{2}}}{(x - \sqrt{10}) (x + \sqrt{10})}]$

= $\lim_{x \to \sqrt{10}} [\frac{\sqrt{7 + 2 x} - \sqrt{5 + 2 + 2 \sqrt{5} \sqrt{2}}}{(x - \sqrt{10}) (x + \sqrt{10})}]$

= $\lim_{x \to \sqrt{10}} [\frac{\sqrt{7 + 2 x} - \sqrt{7 + 2 \sqrt{10}}}{(x - \sqrt{10}) (x + \sqrt{10})}]$

Rationalising the numerator:

$\lim_{x \to \sqrt{10}} [\frac{(\sqrt{7 + 2 x} - \sqrt{7 + 2 \sqrt{10}}) (\sqrt{7 + 2 x} + \sqrt{7 + 2 \sqrt{10}})}{(x - \sqrt{10}) (x + \sqrt{10}) (\sqrt{7 + 2 x} + \sqrt{7 + 2 \sqrt{10}})}]$

= $\lim_{x \to \sqrt{10}} [\frac{(7 + 2 x) - (7 + 2 \sqrt{10})}{(x - \sqrt{10}) (x + \sqrt{10}) (\sqrt{7 + 2 x} + \sqrt{7 + 2 \sqrt{10}})}]$

= $\lim_{x \to \sqrt{10}} [\frac{2 (x - \sqrt{10})}{(x - \sqrt{10}) (x + \sqrt{10}) (\sqrt{7 + 2 x} + \sqrt{7 + 2 \sqrt{10}})}]$

= $\frac{2}{(\sqrt{10} + \sqrt{10}) (2 \sqrt{7 + 2 \sqrt{10}})}$

$= \frac{2}{(2 \sqrt{10}) \times 2 \sqrt{7 + 2 \sqrt{10}}} = \frac{1}{2 \sqrt{10} \sqrt{{(\sqrt{5} + \sqrt{2})}^{2}}}$

= $\frac{1}{2 \sqrt{10} (\sqrt{5} + \sqrt{2})} \times \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}}$

$= \frac{1}{2 \sqrt{10}} [\frac{\sqrt{5} - \sqrt{2}}{{(\sqrt{5})}^{2} - {(\sqrt{2})}^{2}}]$

= $\frac{\sqrt{5} - \sqrt{2}}{6 \sqrt{10}}$

Page No 29.29:

Question 33:

$\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2 x} - (\sqrt{3} + \sqrt{2})}{x^{2} - 6}$

Answer:

$\lim_{x \to \sqrt{6}} [\frac{\sqrt{5 + 2 x} - (\sqrt{3} + \sqrt{2})}{x^{2} - 6}]$

= $\lim_{x \to \sqrt{6}} [\frac{\sqrt{5 + 2 x} - \sqrt{{(\sqrt{3} + \sqrt{2})}^{2}}}{x^{2} - {(\sqrt{6})}^{2}}]$

= $\lim_{x \to \sqrt{6}} [\frac{\sqrt{5 + 2 x} - \sqrt{3 + 2 + 2 \sqrt{6}}}{(x - \sqrt{6}) (x + \sqrt{6})}]$

= $\lim_{x \to \sqrt{6}} [\frac{\sqrt{5 + 2 x} - \sqrt{5 + 2 \sqrt{6}}}{(x - \sqrt{6}) (x + \sqrt{6})}]$

Rationalising the numerator:

$\lim_{x \to \sqrt{6}} [\frac{(\sqrt{5 + 2 x} - \sqrt{5 + 2 \sqrt{6}}) (\sqrt{5 + 2 x} + \sqrt{5 + 2 \sqrt{6}})}{(x - \sqrt{6}) (x + \sqrt{6}) (\sqrt{5 + 2 x} + \sqrt{5 + 2 \sqrt{6}})}]$

= $\lim_{x \to \sqrt{6}} [\frac{(5 + 2 x) - (5 + 2 \sqrt{6})}{(x - \sqrt{6}) (x + \sqrt{6}) (\sqrt{5 + 2 x} + \sqrt{5 + 2 \sqrt{6}})}]$

= $\lim_{x \to \sqrt{6}} [\frac{2 (x - \sqrt{6})}{(x - \sqrt{6}) (x + \sqrt{6}) (\sqrt{5 + 2 x} + \sqrt{5 + 2 \sqrt{6}})}]$

= $\frac{2}{(\sqrt{6} + \sqrt{6}) (\sqrt{5 + 2 \sqrt{6}} + \sqrt{5 + 2 \sqrt{6}})}$

= $\frac{1}{2 \sqrt{6} (\sqrt{{(\sqrt{3} + \sqrt{2})}^{2}})}$

= $\frac{1}{2 \sqrt{6} (\sqrt{3} + \sqrt{2})}$

= $\frac{1}{2 \sqrt{6} (\sqrt{3} + \sqrt{2})} \times \frac{(\sqrt{3} - \sqrt{2})}{(\sqrt{3} - \sqrt{2})}$

= $\frac{\sqrt{3} - \sqrt{2}}{2 \sqrt{6} (3 - 2)}$

= $\frac{\sqrt{3} - \sqrt{2}}{2 \sqrt{6}}$

Page No 29.29:

Question 34:

$\lim_{x \to \sqrt{2}} \frac{\sqrt{3 + 2 x} - (\sqrt{2} + 1)}{x^{2} - 2}$

Answer:

$\lim_{x \to \sqrt{2}} [\frac{\sqrt{3 + 2 x} - (\sqrt{2} + 1)}{x^{2} - 2}]$

= $\lim_{x \to \sqrt{2}} [\frac{\sqrt{3 + 2 x} - \sqrt{{(\sqrt{2} + 1)}^{2}}}{(x - \sqrt{2}) (x + \sqrt{2})}]$

= $\lim_{x \to \sqrt{2}} [\frac{\sqrt{3 + 2 x} - \sqrt{2 + 1 + 2 \sqrt{2}}}{(x - \sqrt{2}) (x + \sqrt{2})}]$

= $\lim_{x \to \sqrt{2}} [\frac{(\sqrt{3 + 2 x} - \sqrt{3 + 2 \sqrt{2}})}{(x - \sqrt{2}) (x + \sqrt{2})}]$

Rationalising the numerator:

$\lim_{x \to \sqrt{2}} [\frac{(\sqrt{3 + 2 x} - \sqrt{3 + 2 \sqrt{2}}) (\sqrt{3 + 2 x} + \sqrt{3 + 2 \sqrt{2}})}{(x - \sqrt{2}) (x + \sqrt{2}) (\sqrt{3 + 2 x} + \sqrt{3 + 2 \sqrt{2}})}]$

= $\lim_{x \to \sqrt{2}} [\frac{(3 + 2 x) - (3 + 2 \sqrt{2})}{(x - \sqrt{2}) (x + \sqrt{2}) (\sqrt{3 + 2 x} + \sqrt{3 + 2 \sqrt{2}})}]$

= $\lim_{x \to \sqrt{2}} [\frac{2 (x - \sqrt{2})}{(x - \sqrt{2}) (x + \sqrt{2}) (\sqrt{3 + 2 x} + \sqrt{3 + 2 \sqrt{2}})}]$

= $\frac{2}{(\sqrt{2} + \sqrt{2}) (\sqrt{3 + 2 \sqrt{2}} + \sqrt{3 + 2 \sqrt{2}})}$

= $\frac{2}{(2 \sqrt{2}) (2 \sqrt{3 + 2 \sqrt{2}})}$

= $\frac{1}{2 \sqrt{2} (\sqrt{3 + 2 \sqrt{2}})}$

= $\frac{1}{2 \sqrt{2} \sqrt{{(\sqrt{2} + 1)}^{2}}}$

= $\frac{1}{2 \sqrt{2} (\sqrt{2} + 1)} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1}$

= $\frac{\sqrt{2} - 1}{2 \sqrt{2} (2 - 1)}$

= $\frac{\sqrt{2} - 1}{2 \sqrt{2}}$

Page No 29.33:

Question 1:

$\lim_{x \to a} \frac{{(x + 2)}^{5 / 2} - {(a + 2)}^{5 / 2}}{x - a}$

Answer:

$\lim_{x \to a} [\frac{{(x + 2)}^{\frac{5}{2}} - {(a + 2)}^{\frac{5}{2}}}{x - a}] = \lim_{x \to a} [\frac{{(x + 2)}^{\frac{5}{2}} - {(a + 2)}^{\frac{5}{2}}}{(x + 2) - (a + 2)}]$

Let y = x + 2 and b = a + 2.

When x → a, then x + 2 → a + 2.
⇒ y → b

$\lim_{y \to b} [\frac{y^{\frac{5}{2}} - b^{\frac{5}{2}}}{y - b}] = \frac{5}{2} {(b)}^{\frac{5}{2} - 1} = \frac{5}{2} b^{\frac{3}{2}} = \frac{5}{2} {(a + 2)}^{\frac{3}{2}}$

Page No 29.33:

Question 2:

$\lim_{x \to a} \frac{{(x + 2)}^{3 / 2} - {(a + 2)}^{3 / 2}}{x - a}$

Answer:

$\lim_{x \to a} = [\frac{{(x + 2)}^{\frac{3}{2}} - {(a + 2)}^{\frac{3}{2}}}{x - a}] = \lim_{x \to a} [\frac{{(x + 2)}^{\frac{3}{2}} - {(a + 2)}^{\frac{3}{2}}}{(x + 2) - (a + 2)}]$

Let y = x + 2 and b = a + 2.

When x → a and x + 2 → a + 2.
$\Rightarrow$ y → b

$\lim_{y \to b} [\frac{y^{\frac{3}{2}} - b^{\frac{3}{2}}}{y - b}] = \frac{3}{2} {(b)}^{\frac{3}{2} - 1} = \frac{3}{2} b^{\frac{1}{2}} = \frac{3}{2} {(a + 2)}^{\frac{1}{2}}$

Page No 29.33:

Question 3:

$\lim_{x \to 0} \frac{{(1 + x)}^{6} - 1}{{(1 + x)}^{2} - 1}$

Answer:

$\lim_{x \to 0} [\frac{{(1 + x)}^{6} - 1}{{(1 + x)}^{2} - 1}] = \lim_{x \to 0} [\frac{{(1 + x)}^{6} - 1}{x} \times \frac{x}{{(1 + x)}^{2} - 1}] = \lim_{x \to 0} [\frac{{(1 + x)}^{6} - 1^{6}}{(1 + x) - 1} \times \frac{(1 + x) - 1}{{(1 + x)}^{2} - 1}]$

Let y = 1 + x

When x → 0, then 1 + x → 1.
$\Rightarrow$ y → 1

$\lim_{y \to 1} [(\frac{y^{6} - 1^{6}}{y - 1}) \times \frac{(y - 1)}{y^{2} - 1^{2}}] = \frac{6 \times {(1)}^{6 - 1}}{2 \times {(1)}^{2 - 1}} = \frac{6}{2} = 3$

Page No 29.33:

Question 4:

$\lim_{x \to a} \frac{x^{2 / 7} - a^{2 / 7}}{x - a}$

Answer:

$\lim_{x \to a} [\frac{x^{\frac{2}{7}} - a^{\frac{2}{7}}}{x - a}] = \frac{2}{7} a^{\frac{2}{7} - 1} [∵ \lim_{x \to a} [\frac{x^{n} - a^{n}}{x - a}] = n a^{n - 1}] = \frac{2}{7} a^{- \frac{5}{7}}$

Page No 29.33:

Question 5:

$\lim_{x \to a} \frac{x^{5 / 7} - a^{5 / 7}}{x^{2 / 7} - a^{2 / 7}}$

Answer:

$\lim_{x \to a} [\frac{x^{\frac{5}{7}} - a^{\frac{5}{7}}}{x^{\frac{2}{7}} - a^{\frac{2}{7}}}] = \lim_{x \to a} [(\frac{x^{\frac{5}{7}} - a^{\frac{5}{7}}}{x - a}) \times (\frac{x - a}{x^{\frac{2}{7}} - a^{\frac{2}{7}}})] = \frac{5}{7} a^{\frac{5}{7} - 1} \times \frac{1}{\frac{2}{7} a^{\frac{2}{7} - 1}} = \frac{5}{2} a^{- \frac{2}{7}} a^{\frac{5}{7}} = \frac{5}{2} a^{- \frac{2}{7} + \frac{5}{7}} = \frac{5}{2} a^{\frac{3}{7}}$

Page No 29.33:

Question 6:

$\lim_{x \to - 1 / 2} \frac{8 x^{3} + 1}{2 x + 1}$

Answer:

$\lim_{x \to - \frac{1}{2}} [\frac{8 x^{3} + 1}{2 x + 1}] = \lim_{x \to - \frac{1}{2}} [\frac{{(2 x)}^{3} - (- 1)}{2 x - (- 1)}] When x \to - \frac{1}{2}, then 2x → –1 .$
Let y = 2x

$\lim_{y \to - 1} [\frac{y^{3} - {(- 1)}^{3}}{y - (- 1)}] = 3 {(- 1)}^{3 - 1} = 3 \times {(- 1)}^{2} = 3$

Page No 29.33:

Question 7:

$\lim_{x \to 27} \frac{(x^{1 / 3} + 3) (x^{1 / 3} - 3)}{x - 27}$

Answer:

$\lim_{x \to 27} \frac{[x^{\frac{1}{3}} + 3] [x^{\frac{1}{3}} - 3]}{x - 27} = \lim_{x \to 27} [\frac{(x^{\frac{1}{3}} + 3) (x^{\frac{1}{3}} - 3)}{{(x^{\frac{1}{3}})}^{3} - 3^{3}}] x \to 27 ∴ x^{\frac{1}{3}} \to 3 Let y = x^{\frac{1}{3}} \lim_{y \to 3} [\frac{(y + 3) (y - 3)}{y^{3} - 3^{3}}] = \frac{(3 + 3)}{3 \times 3^{3 - 1}} = \frac{6}{3 \times 9} = \frac{2}{9}$

Page No 29.33:

Question 8:

$\lim_{x \to 4} \frac{x^{3} - 64}{x^{2} - 16}$

Answer:

$\lim_{x \to 4} [\frac{x^{3} - 64}{x^{2} - 16}] = \lim_{x \to 4} [\frac{x^{3} - 64}{x - 4} \times \frac{x - 4}{x^{2} - 16}] = \lim_{x \to 4} [\frac{x^{3} - 4^{3}}{x - 4} \times \frac{x - 4}{x^{2} - 4^{2}}] = 3 {(4)}^{3 - 1} \times \frac{1}{2 {(4)}^{2 - 1}} = \frac{3 \times 16}{2 \times 4} = 6$

Page No 29.33:

Question 9:

$\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}$

Answer:

$\lim_{x \to 1} [\frac{x^{15} - 1}{x^{10} - 1}] = \lim_{x \to 1} [\frac{x^{15} - {(1)}^{15}}{x - 1} \times \frac{x - 1}{x^{10} - {(1)}^{10}}] = \frac{15 {(1)}^{15 - 1}}{10 {(1)}^{10 - 1}} = \frac{3}{2}$

Page No 29.33:

Question 10:

$\lim_{x \to - 1} \frac{x^{3} + 1}{x + 1}$

Answer:

$\lim_{x \to - 1} [\frac{x^{3} + 1}{x + 1}] = \lim_{x \to - 1} [\frac{x^{3} - (- 1)}{x - (- 1)}] = \lim_{x \to - 1} [\frac{x^{3} - {(- 1)}^{3}}{x - (- 1)}] = 3 {(- 1)}^{3 - 1} = 3$

Page No 29.33:

Question 11:

$\lim_{x \to a} \frac{x^{2 / 3} - a^{2 / 3}}{x^{3 / 4} - a^{3 / 4}}$

Answer:

$\lim_{x \to a} [\frac{x^{\frac{2}{3}} - a^{\frac{2}{3}}}{x^{\frac{3}{4}} - a^{\frac{3}{4}}}] = \lim_{a \to a} [(\frac{x^{\frac{2}{3}} - a^{\frac{2}{3}}}{x - a}) \times (\frac{x - a}{x^{\frac{3}{4}} - a^{\frac{3}{4}}})] = \frac{2}{3} {(a)}^{\frac{2}{3} - 1} \times \frac{1}{\frac{3}{4} a^{\frac{3}{4} - 1}} = \frac{8}{9} \frac{a^{- \frac{1}{3}}}{a^{- \frac{1}{4}}} = \frac{8}{9} a^{- \frac{1}{3} + \frac{1}{4}} = \frac{8}{9} a^{- \frac{1}{12}}$

Page No 29.33:

Question 12:

If $\lim_{x \to 3} \frac{x^{n} - 3^{n}}{x - 3} = 108,$ find the value of n.

Answer:

$\lim_{x \to 3} [\frac{x^{n} - 3^{n}}{x - 3}] = 108$

⇒ x(3)^{n – 1} = 108
⇒ x(3)^{n – 1} = 4 × 3³

On comparing LHS and RHS, we observe that x is equal to 4.

$\begin{matrix} 2 & 108 \\ 2 & 54 \\ 3 & 27 \\ 3 & 9 \\ 3 & 3 \\ 1 \end{matrix}$

Page No 29.33:

Question 13:

If $\lim_{x \to a} \frac{x^{9} - a^{9}}{x - a} = 9,$ find all possible values of a.

Answer:

$\lim_{x \to a} [\frac{x^{9} - a^{9}}{x - a}] = 9 \Rightarrow 9 a^{8} = 9 \Rightarrow a^{8} = 1 \Rightarrow a = \pm 1$

Page No 29.33:

Question 14:

If $\lim_{x \to a} \frac{x^{5} - a^{5}}{x - a} = 405,$ find all possible values of a.

Answer:

$\lim_{x \to a} [\frac{x^{5} - a^{5}}{x - a}] = 405 \Rightarrow 5 a^{4} = 405 \Rightarrow a^{4} = 81 \Rightarrow a^{2} = 9 \Rightarrow a = \pm 3$

Page No 29.33:

Question 15:

If $\lim_{x \to a} \frac{x^{9} - a^{9}}{x - a} = \lim_{x \to 5} (4 + x),$ find all possible values of a.

Answer:

$\lim_{x \to a} [\frac{x^{9} - a^{9}}{x - a}] = \lim_{x \to 5} (4 + x) \Rightarrow 9 a^{9 + 1} = 9 \Rightarrow a^{8} = 1 \Rightarrow a = \pm 1$

Page No 29.33:

Question 16:

If $\lim_{x \to a} \frac{x^{3} - a^{3}}{x - a} = \lim_{x \to 1} \frac{x^{4} - 1}{x - 1},$ find all possible values of a.

Answer:

$\lim_{x \to a} [\frac{x^{3} - a^{3}}{x - a}] = \lim_{x \to 1} [\frac{x^{4} - 1^{4}}{x - 1}] \Rightarrow 3 a^{3 - 1} = 4 {(1)}^{4 - 1} \Rightarrow 3 a^{2} = 4 \Rightarrow a^{2} = \frac{4}{3} \Rightarrow a = \pm \frac{2}{\sqrt{3}}$

Page No 29.38:

Question 1:

$\lim_{x \to \infty} \frac{(3 x - 1) (4 x - 2)}{(x + 8) (x - 1)}$

Answer:

$\lim_{x \to \infty} \frac{(3 x - 1) (4 x - 2)}{(x + 8) (x - 1)} Dividing the numerator and the denominator by x^{2} : \lim_{x \to \infty} \frac{(\frac{3 x - 1}{x}) (\frac{4 x - 2}{x})}{(\frac{x + 8}{x}) (\frac{x - 1}{x})} = \lim_{x \to \infty} [\frac{(3 - \frac{1}{x}) (4 - \frac{2}{x})}{(1 + \frac{8}{x}) (1 - \frac{1}{x})}] When x \to \infty, then \frac{1}{x} \to 0 . \frac{3 \times 4}{1} = 12$

Page No 29.38:

Question 2:

$\lim_{x \to \infty} \frac{3 x^{3} - 4 x^{2} + 6 x - 1}{2 x^{3} + x^{2} - 5 x + 7}$

Answer:

$\lim_{x \to \infty} [\frac{3 x^{3} - 4 x^{2} + 6 x - 1}{2 x^{3} + x^{2} - 5 x + 7}] Dividing the numerator and the denominator by x^{3} : \lim_{x \to \infty} [\frac{3 - \frac{4}{x} + \frac{6}{x^{2}} - \frac{1}{x^{3}}}{2 + \frac{1}{x} - \frac{5}{x^{2}} + \frac{7}{x^{3}}}] When x \to \infty, then \frac{1}{x}, \frac{1}{x^{2}}, \frac{1}{x^{3}} \to 0 . \Rightarrow \frac{3}{2}$

Page No 29.38:

Question 3:

$\lim_{x \to \infty} \frac{5 x^{3} - 6}{\sqrt{9 + 4 x^{6}}}$

Answer:

$\lim_{x \to \infty} [\frac{5 x^{3} - 6}{\sqrt{9 + 4 x^{6}}}] Dividing the numerator and the denominator by x : \lim_{x \to \infty} [\frac{\frac{5 x^{3} - 6}{x^{3}}}{\frac{\sqrt{9 + 4 x^{6}}}{x^{3}}}] = \lim_{x \to \infty} [\frac{5 - \frac{6}{x^{3}}}{\sqrt{\frac{9}{x^{6}} + 4}}] = \frac{5}{\sqrt{4}} = \frac{5}{2}$

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Question 4:

$\lim_{x \to \infty} \sqrt{x^{2} + c x - x}$

Answer:

$\lim_{x \to \infty} (\sqrt{x^{2} + c x - x})$

It is of the form ∞ - ∞.

Rationalising the numerator:

$\lim_{x \to \infty} (\sqrt{x^{2} + c x} - x) \frac{(\sqrt{x^{2} + c x} + x)}{(\sqrt{x^{2} + c x} + x)} = \lim_{x \to \infty} (\frac{x^{2} + c x - x^{2}}{\sqrt{x^{2} + c x} + x}) Dividing the numerator and the denomiator by x : \lim_{x \to \infty} [\frac{\frac{c x}{x}}{\frac{\sqrt{x^{2} + c x} + x}{x}}] = \lim_{x \to \infty} [\frac{c}{\sqrt{1 + \frac{c}{x}} + 1}] When x \to \infty, then \frac{1}{x} \to 0 . \Rightarrow \frac{c}{\sqrt{1} + 1} = \frac{c}{2}$

Page No 29.38:

Question 5:

$\lim_{x \to \infty} \sqrt{x + 1} - \sqrt{x}$

Answer:

$\lim_{x \to \infty} (\sqrt{x + 1} - \sqrt{x})$

It is of the form ∞–∞.

On rationalising, we get:
$\lim_{x \to \infty} (\sqrt{x + 1} - \sqrt{x}) \times (\frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}}) = \lim_{x \to \infty} (\frac{x + 1 - x}{(\sqrt{x + 1} + \sqrt{x})}) = \frac{1}{\infty} = 0$

Page No 29.38:

Question 6:

$\lim_{x \to \infty} \sqrt{x^{2} + 7 x - x}$

Answer:

$\lim_{x \to \infty} [\sqrt{x^{2} + 7 x} - x] Rationalising the numerator : \lim_{x \to \infty} [(\sqrt{x^{2} + 7 x} - x) \frac{(\sqrt{x^{2} + 7 x} + x)}{(\sqrt{x^{2} + 7 x} + x)}] = \lim_{x \to \infty} [\frac{x^{2} + 7 x - x^{2}}{(\sqrt{x^{2} + 7 x} + x)}] Dividing the numerator and the denominator by x : \lim_{x \to \infty} [\frac{7}{\frac{\sqrt{x^{2} + 7 x}}{x} + 1}] = \lim_{x \to \infty} [\frac{7}{\frac{\sqrt{x^{2} + 7 x}}{x} + 1}] When x \to \infty, then \frac{1}{x} \to 0 . \Rightarrow \frac{7}{\sqrt{1} + 1} = \frac{7}{2}$

Page No 29.38:

Question 7:

$\lim_{x \to \infty} \frac{x}{\sqrt{4 x^{2} + 1} - 1}$

Answer:

$\lim_{x \to \infty} [\frac{x}{\sqrt{4 x^{2} + 1} - 1}] Rationalising the denominator : \lim_{x \to \infty} [\frac{x}{(\sqrt{4 x^{2} + 1} - 1)} \frac{(\sqrt{4 x^{2} + 1} + 1)}{(\sqrt{4 x^{2} + 1} + 1)}] = \lim_{x \to \infty} [\frac{x (\sqrt{4 x^{2} + 1} + 1)}{4 x^{2} + 1 - 1}] = \lim_{x \to \infty} [\frac{\sqrt{4 x^{2} + 1} + 1}{4 x}] Dividing the numerator and the denominator by x : \lim_{x \to \infty} [\frac{\frac{\sqrt{4 x^{2} + 1}}{x} + \frac{1}{x}}{4}] = \lim_{x \to \infty} [\frac{\sqrt{\frac{4 x^{2} + 1}{x^{2}}} + \frac{1}{x}}{4}] = \lim_{x \to \infty} [\frac{\sqrt{4 + \frac{1}{x^{2}}} + \frac{1}{x}}{4}] x \to \infty ∴ \frac{1}{x}, \frac{1}{x^{2}} \to 0 = \frac{\sqrt{4}}{4} = \frac{2}{4} = \frac{1}{2}$

Page No 29.38:

Question 8:

$\lim_{n \to \infty} \frac{n^{2}}{1 + 2 + 3 + . . . + n}$

Answer:

$\lim_{n \to \infty} [\frac{n^{2}}{1 + 2 + 3 . . . . . n}] It is of the form \frac{\infty}{\infty} . \Rightarrow \lim_{n \to \infty} [\frac{n^{2}}{n \frac{(n + 1)}{2}}] = \lim_{n \to \infty} [\frac{2 n}{n + 1}] Dividing the numerator and the denominator by n: \lim_{n \to \infty} \frac{2}{1 + \frac{1}{n}} = 2$

Page No 29.39:

Question 9:

$\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}$

Answer:

$\lim_{x \to \infty} [\frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}] = \lim_{x \to \infty} [\frac{\frac{3}{x} + \frac{4}{x^{2}}}{\frac{5}{x} + \frac{6}{x^{2}}}] = \lim_{x \to \infty} [\frac{3 x + 4}{5 x + 6}] Dividing the numerator and the denominator by x: \lim_{x \to \infty} [\frac{3 + \frac{4}{x}}{5 + \frac{6}{x}}] = \frac{3}{5}$

Page No 29.39:

Question 10:

$\lim_{x \to \infty} \frac{\sqrt{x^{2} + a^{2}} - \sqrt{x^{2} + b^{2}}}{\sqrt{x^{2} + c^{2}} - \sqrt{x^{2} + d^{2}}}$

Answer:

$\lim_{x \to \infty} [\frac{\sqrt{x^{2} + a^{2}} - \sqrt{x^{2} + b^{2}}}{\sqrt{x^{2} + c^{2}} - \sqrt{x^{2} + d^{2}}}] Rationalising the numerator and the denominator: \lim_{x \to \infty} [\frac{(\sqrt{x^{2} + a^{2}} - \sqrt{x^{2} + b^{2}})}{(\sqrt{x^{2} + c^{2}} - \sqrt{x^{2} + d^{2}})} \times \frac{(\sqrt{x^{2} + c^{2}} + \sqrt{x^{2} + d^{2}})}{(\sqrt{x^{2} + c^{2}} + \sqrt{x^{2} + d^{2}})} \times \frac{(\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} + b^{2}})}{(\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} + b^{2}})}] = \lim_{x \to \infty} [\frac{(\sqrt{x^{2} + a^{2}} - \sqrt{x^{2} + b^{2}}) (\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} + b^{2}}) (\sqrt{x^{2} + c^{2}} + \sqrt{x^{2} + d^{2}})}{(\sqrt{x^{2} + c^{2}} - \sqrt{x^{2} + d^{2}}) (\sqrt{x^{2} + c^{2}} + \sqrt{x^{2} + d^{2}}) (\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} + b^{2}})}] = \lim_{x \to \infty} \frac{(x^{2} + a^{2}) - (x^{2} + b^{2})}{(x^{2} + c^{2}) - (x^{2} + d^{2})} \times (\frac{\sqrt{x^{2} + c^{2}} + \sqrt{x^{2} + d^{2}}}{\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} + b^{2}}}) \underset{x \to \infty}{= \lim} (\frac{a^{2} - b^{2}}{c^{2} - d^{2}}) (\frac{\sqrt{x^{2} + c^{2}} + \sqrt{x^{2} + d^{2}}}{\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} + b^{2}}}) Dividing the numerator and the denominator by x : \lim_{x \to \infty} (\frac{a^{2} - b^{2}}{c^{2} - d^{2}}) (\frac{\sqrt{1 + \frac{c^{2}}{x^{2}}} + \sqrt{1 + \frac{d^{2}}{x^{2}}}}{\sqrt{1 + \frac{1}{x^{2}}} + \sqrt{1 + \frac{b^{2}}{x^{2}}}}) As x \to \infty, \frac{1}{x}, \frac{1}{x^{2}} \to 0 = (\frac{a^{2} - b^{2}}{c^{2} - d^{2}}) (\frac{\sqrt{1} + \sqrt{1}}{\sqrt{1} + \sqrt{1}}) = \frac{a^{2} - b^{2}}{c^{2} - d^{2}}$

Page No 29.39:

Question 11:

$\lim_{n \to \infty} [\frac{(n + 2)! + (n + 1)!}{(n + 2)! - (n + 1)!}]$

Answer:

$\lim_{n \to \infty} [\frac{(n + 2)! + (n + 1)!}{(n + 2)! - (n + 1)!}] = \lim_{n \to \infty} [\frac{(n + 2) (n + 1)! + (n + 1)!}{(n + 2) (n + 1)! - (n + 1)!}] = \lim_{n \to \infty} [\frac{(n + 1)!}{(n + 1)!} \times \frac{(n + 2 + 1)}{(n + 2 - 1)}] = \lim_{n \to \infty} [\frac{n + 3}{n + 1}]$

Dividing the numerator and the denominator by n:

$\lim_{n \to \infty} [\frac{1 + \frac{3}{n}}{1 + \frac{1}{n}}] When n \to \infty, then \frac{1}{n} \to 0 . \Rightarrow \frac{1}{1} = 1$

Page No 29.39:

Question 12:

$\lim_{x \to \infty} [x \{\sqrt{x^{2} + 1} - \sqrt{x^{2} - 1}\}]$

Answer:

$\lim_{x \to \infty} [x \{\sqrt{x^{2} + 1} - \sqrt{x^{2} - 1}\}] Rationalising the numerator : \lim_{x \to \infty} [x \{\frac{(\sqrt{x^{2} + 1} - \sqrt{x^{2} - 1}) (\sqrt{x^{2} + 1} + \sqrt{x^{2} - 1})}{(\sqrt{x^{2} + 1} + \sqrt{x^{2} - 1})}\}] = \lim_{x \to \infty} [x \{\frac{(x^{2} + 1) - (x^{2} - 1)}{(\sqrt{x^{2} + 1} + \sqrt{x^{2} - 1})}\}] = \lim_{x \to \infty} [\frac{x \times 2}{(\sqrt{x^{2} + 1} + \sqrt{x^{2} - 1})}] Dividing the numerator and the denominator by x : \lim_{x \to \infty} [\frac{2}{\sqrt{\frac{x^{2} + 1}{x^{2}}} + \sqrt{\frac{x^{2} - 1}{x^{2}}}}] = \lim_{x \to \infty} [\frac{2}{\sqrt{\frac{x^{2} + 1}{x^{2}}} + \sqrt{\frac{x^{2} - 1}{x^{2}}}}] = \lim_{x \to \infty} [\frac{2}{\sqrt{1 + \frac{1}{x^{2}}} + \sqrt{1 - \frac{1}{x^{2}}}}] = \frac{2}{\sqrt{1} + \sqrt{1}} = 2$

Page No 29.39:

Question 13:

$\lim_{x \to \infty} [\{\sqrt{x + 1} - \sqrt{x}\} \sqrt{x + 2}]$

Answer:

$\lim_{x \to \infty} [\{\sqrt{x + 1} - \sqrt{x}\} \sqrt{x + 2}] Rationalising the numerator : \lim_{x \to \infty} [\frac{(\sqrt{x + 2}) \{\sqrt{x + 1} - \sqrt{x}\} \{\sqrt{x + 1} + \sqrt{x}\}}{(\sqrt{x + 1} + \sqrt{x})}] \Rightarrow \lim_{x \to \infty} [\frac{(\sqrt{x + 2}) (x + 1 - x)}{(\sqrt{x + 1} + \sqrt{x})}] Dividing the numerator and the denominator by \sqrt{x} : \lim_{x \to \infty} [\frac{\frac{\sqrt{x + 2}}{\sqrt{x}}}{\frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x}}}] = \lim_{x \to \infty} [\frac{\sqrt{1 + \frac{2}{x}}}{\sqrt{1 + \frac{1}{x}} + 1}] = \frac{1}{\sqrt{1} + 1} = \frac{1}{2}$

Page No 29.39:

Question 14:

$\lim_{n \to \infty} [\frac{1^{2} + 2^{2} + . . . + n^{2}}{n^{3}}]$

Answer:

$= \lim_{n \to \infty} [\frac{1^{2} + 2^{2} + . . . + n^{2}}{n^{3}}] = \lim_{n \to \infty} [\frac{n (n + 1) (2 n + 1)}{6 n^{3}}] = \lim_{n \to \infty} [\frac{(n + 1) (2 n + 1)}{6 n^{2}}] = \lim_{n \to \infty} [(\frac{n + 1}{n}) (\frac{2 n + 1}{n}) \times \frac{1}{6}] = \lim_{n \to \infty} [(1 + \frac{1}{n}) (2 + \frac{1}{n}) \times \frac{1}{6}] = \frac{2}{6} = \frac{1}{3}$

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Question 15:

$\lim_{n \to \infty} [\frac{1 + 2 + 3 . . . . . . n - 1}{n^{2}}]$

Answer:

$\Rightarrow \lim_{n \to \infty} (\frac{1 + 2 + 3 + . . . n - 1}{n^{2}}) \Rightarrow \lim_{n \to \infty} [\frac{n (n - 1)}{2 n^{2}}] \Rightarrow \lim_{n \to \infty} [(1 - \frac{1}{n}) \times \frac{1}{2}] When n \to \infty, then \frac{1}{n} \to 0 . = \frac{1}{2}$

Page No 29.39:

Question 16:

$\lim_{n \to \infty} [\frac{1^{3} + 2^{3} + . . . . n^{3}}{n^{4}}]$

Answer:

$\lim_{n \to \infty} [\frac{1^{3} + 2^{3} + . . . + n^{3}}{n^{4}}] = \lim_{n \to \infty} [\frac{{[\frac{n (n + 1)}{2}]}^{2}}{n^{4}}] = \lim_{n \to \infty} [\frac{n^{2} {(n + 1)}^{2}}{4 n^{4}}] = \lim_{n \to \infty} [\frac{n^{2} {(n + 1)}^{2}}{4 n^{4}}] = \lim_{n \to \infty} [\frac{{(n + 1)}^{2}}{4 n^{2}}] = \lim_{n \to \infty} [{(\frac{n + 1}{n})}^{2} \times \frac{1}{4}] = \lim_{n \to \infty} [{(1 + \frac{1}{n})}^{2} \times \frac{1}{4}] When n \to \infty, then \frac{1}{n} \to 0 . \Rightarrow \frac{1}{4}$

Page No 29.39:

Question 17:

$\lim_{n \to \infty} [\frac{1^{3} + 2^{3} + . . . n^{3}}{{(n - 1)}^{4}}]$

Answer:

$\lim_{n \to \infty} [\frac{1^{3} + 2^{3} + . . . + n^{3}}{{(n - 1)}^{4}}] = \lim_{n \to \infty} [\frac{{[\frac{n (n + 1)}{2}]}^{2}}{{(n - 1)}^{4}}] = \lim_{n \to \infty} [\frac{n^{2} {(n + 1)}^{2}}{4 {(n - 1)}^{4}}]$

Dividing the numerator and the denominator by n⁴:

$\lim_{n \to \infty} [\frac{\frac{n^{2} {(n + 1)}^{2}}{n^{4}}}{4 \frac{{(n - 1)}^{4}}{n^{4}}}] = \lim_{n \to \infty} [\frac{\frac{{(n + 1)}^{2}}{n^{2}}}{4 {(\frac{n - 1}{n})}^{4}}] = \lim_{n \to \infty} [\frac{{(1 + \frac{1}{n})}^{2}}{4 {(1 - \frac{1}{n})}^{4}}] When n \to \infty, then \frac{1}{n} \to 0 . \frac{{(1 + 0)}^{2}}{4 {(1 - 0)}^{4}} = \frac{1}{4}$

Page No 29.39:

Question 18:

$\lim_{x \to \infty} [\sqrt{x} \{\sqrt{x + 1} - \sqrt{x}\}]$

Answer:

$\lim_{x \to \infty} [\sqrt{x} \{\sqrt{x + 1} - \sqrt{x}\}] = \lim_{x \to \infty} [\sqrt{x} \{(\sqrt{x + 1} - \sqrt{x}) \frac{\sqrt{x + 1} + \sqrt{x}}{(\sqrt{x + 1} + \sqrt{x})}\}] = \lim_{x \to \infty} [\frac{\sqrt{x} \{(x + 1) - x\}}{(\sqrt{x + 1} + \sqrt{x})}]$

Dividing the numerator and the denominator by $\sqrt{x}$ :

$\lim_{x \to \infty} [\frac{1}{\frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x}}}] = \lim_{x \to \infty} [\frac{1}{\sqrt{1 + \frac{1}{x}} + 1}] As x \to \infty, \frac{1}{x} \to 0 = \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{2}$

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Question 19:

$\lim_{n \to \infty} [\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + . . . + \frac{1}{3^{n}}]$

Answer:

$\lim_{n \to \infty} [\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + . . . \frac{1}{3^{n}}] = \lim_{n \to \infty} [\frac{1}{3} (1 + \frac{1}{3} + \frac{1}{3^{2}} + . . . . . \frac{1}{3^{n - 1}})] = \lim_{n \to \infty} \frac{[\frac{1}{3} (1 - \frac{1}{3^{n}})]}{1 - \frac{1}{3}} [a + a r + a r^{2} + . . . . . . a r^{n - 1} = a (\frac{1 - r^{n}}{1 - r})] = \lim_{n \to \infty} [\frac{\frac{1}{3} (1 - \frac{1}{3^{n}})}{\frac{2}{3}}] = \lim_{n \to \infty} \frac{1}{2} [(1 - \frac{1}{3^{n}})] As n \to \infty, 3^{n} \to \infty, \frac{1}{3^{n}} \to 0 = \frac{1}{2} (1 - 0) = \frac{1}{2}$

Page No 29.39:

Question 20:

$\lim_{x \to \infty} [\frac{x^{4} + 7 x^{3} + 46 x + a}{x^{4} + 6}]$ , where a is a non-zero real number.

Answer:

$\lim_{x \to \infty} [\frac{x^{4} + 7 x^{3} + 46 x + a}{x^{4} + 6}]$

Dividing the numerator and the denominator by x⁴:

$\lim_{x \to \infty} [\frac{1 + \frac{7}{x} + \frac{46}{x^{2}} + \frac{9}{x^{4}}}{1 + \frac{6}{x^{4}}}] As x \to \infty, \frac{1}{x}, \frac{1}{x^{2}}, \frac{1}{x^{3}}, \frac{1}{x^{4}} \to 0 = \frac{1}{1}$

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Question 21:

$f (x) = \frac{a x^{2} + b}{x^{2} + 1}, \lim_{x \to 0} f (x) = 1$ and $\lim_{x \to \infty} f (x) = 1,$ then prove that f(−2) = f(2) = 1

Answer:

$f (x) = \frac{a x^{2} + b}{x^{2} + 1} \lim_{x \to 0} f (x) = 1 \Rightarrow \lim_{x \to 0} [\frac{a x^{2} + b}{x^{2} + 1}] = 1 \Rightarrow \frac{a \times 0 + b}{a + 1} = 1 \Rightarrow b = 1 Also, \lim_{x \to \infty} f (x) = 1 \lim_{x \to \infty} [\frac{a x^{2} + b}{x^{2} + 1}] = 1$

Dividing the numerator and the denominator by x²:

$\Rightarrow \lim_{x \to \infty} [\frac{a + \frac{b}{x^{2}}}{1 + \frac{1}{x^{2}}}] = 1 As x \to \infty, \frac{1}{x}, \frac{1}{x^{2}} \to 0 \Rightarrow \frac{a + 0}{1 + 0} = 1 \Rightarrow a = 1 ∴ a = 1, b = 1 \Rightarrow f (x) = \frac{x^{2} + 1}{x^{2} + 1} = 1 f (- 2) = 1 [Since f (x) is a constant function, its value does not depend on x .] f (2) = 1$

Page No 29.39:

Question 22:

Show that $\lim_{x \to \infty} (\sqrt{x^{2} + x + 1} - x) \neq \lim_{x \to \infty} (\sqrt{x^{2} + 1} - x)$

Answer:

$\lim_{x \to \infty} (\sqrt{x^{2} + x + 1} - x) \neq \lim_{x \to \infty} (\sqrt{x^{2} + 1} - x) LHS : \lim_{x \to \infty} ((\sqrt{x^{2} + x + 1} - x)) Rationalising the numerator : \lim_{x \to \infty} [\frac{(\sqrt{x^{2} + x + 1} - x) (\sqrt{x^{2} + x + 1} + x)}{(\sqrt{x^{2} + x + 1} + x)}] = \lim_{x \to \infty} [\frac{(x^{2} + x + 1) - x^{2}}{(\sqrt{x^{2} + x + 1} + x)}] = \lim_{x \to \infty} [\frac{x + 1}{(\sqrt{x^{2} + x + 1} + x)}] Dividing the numerator and the denominator by x : \lim_{x \to \infty} [\frac{1 + \frac{1}{x}}{\frac{\sqrt{x^{2} + x + 1}}{x} + 1}] = \lim_{x \to \infty} [\frac{1 + \frac{1}{x}}{\sqrt{\frac{x^{2} + x + 1}{x^{2}}} + 1}] = \lim_{x \to \infty} [\frac{1 + \frac{1}{x}}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^{2}}} + 1}] When x \to \infty, then \frac{1}{x} \to 0 . \frac{1}{\sqrt{1} + 1} = \frac{1}{2} RHS : \lim_{x \to \infty} (\sqrt{x^{2} + 1} - x) [from \infty - \infty]$

Rationalising the numerator:

$\lim_{x \to \infty} [\frac{(\sqrt{x^{2} + 1} - x) (\sqrt{x^{2} + 1} + x)}{(\sqrt{x^{2} + 1} + x)}] = \lim_{x \to \infty} [\frac{x^{2} + 1 - x^{2}}{(\sqrt{x^{2} + 1} + x)}] = \frac{1}{\infty} = 0 ∴ \lim_{x \to \infty} [\sqrt{x^{2} + x + 1} - x] \neq \lim_{x \to \infty} (\sqrt{x^{2} + 1} - x)$

Page No 29.39:

Question 23:

$\lim_{x \to - \infty} (\sqrt{4 x^{2} - 7 x} + 2 x)$

Answer:

$\lim_{x \to - \infty} (\sqrt{4 x^{2} - 7 x} + 2 x)$

Let x = $-$ m
When n → – ∞, then m → ∞.

$\Rightarrow \lim_{m \to \infty} [\sqrt{4 m^{2} + 7 m} - 2 m] = \lim_{m \to \infty} [(\sqrt{4 m^{2} = 7 m} - 2 m) \times \frac{(\sqrt{4 m^{2} + 7 m} + 2 m)}{(\sqrt{4 m^{2} + 7 m} + 2 m)}] = \lim_{m \to \infty} [\frac{(4 m^{2} + 7 m) - {(2 m)}^{2}}{\sqrt{4 m^{2} + 7 m} + 2 m}] = \lim_{m \to \infty} [\frac{4 m^{2} + 7 m - 4 m^{2}}{\sqrt{4 m^{2} + 7 m} + 2 m}]$

Dividing the numerator and the denominator by m:

$\lim_{m \to \infty} [\frac{7}{\sqrt{\frac{4 m^{2} + 7 m}{m^{2}}} + \frac{2 m}{m}}] = \lim_{m \to \infty} [\frac{7}{\sqrt{\frac{4 m^{2}}{m^{2}} + \frac{7 m}{m^{2}}} + 2}] = \lim_{m \to \infty} [\frac{7}{\sqrt{4 + \frac{7}{m}} + 2}] As m \to \infty, \frac{1}{m} \to 0 = \frac{7}{\sqrt{4} + 2} = \frac{7}{4}$

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Question 24:

$\lim_{x \to - \infty} (\sqrt{x^{2} - 8 x} + x)$

Answer:

$\lim_{x \to - \infty} (\sqrt{x^{2} - 8 x} + x)$

Let x = –m

When x → –∞, then m → ∞.

$\lim_{m \to \infty} (\sqrt{m^{2} + 8 m} - m) = \lim_{m \to \infty} (\frac{(\sqrt{m^{2} + 8 m} - m) (\sqrt{m^{2} + 8 m} + m)}{(\sqrt{m^{2} + 8 m} + m)}) = \lim_{m \to \infty} [\frac{m^{2} + 8 m - m^{2}}{\sqrt{m^{2} + 8 m} + m}]$

Dividing the numerator and the denominator by m:

$\lim_{m \to \infty} [\frac{8}{\frac{\sqrt{m^{2} + 8 m} + 1}{m}}] = \lim_{m \to \infty} [\frac{8}{\sqrt{\frac{m^{2}}{m^{2}} + \frac{8 m}{m^{2}}} + 1}] = \lim_{m \to \infty} [\frac{8}{\sqrt{1 + \frac{8}{m}} + 1}] As m \to \infty, \frac{1}{m} \to 0 = \frac{8}{\sqrt{1 + 0} + 1} = 4$

Page No 29.39:

Question 25:

Evaluate: $\lim_{n \to \infty} \frac{1^{4} + 2^{4} + 3^{4} + . . . + n^{4}}{n^{5}} - \lim_{n \to \infty} \frac{1^{3} + 2^{3} + . . . + n^{3}}{n^{5}}$

Answer:

Consider the identity

${(k + 1)}^{5} - k^{5} = 5 k^{4} + 10 k^{3} + 10 k^{2} + 5 k + 1$ .....(1)

Putting k = 1, 2, 3,..., n in (1) and then adding the equations, we have

${(n + 1)}^{5} - 1 = 5 \sum_{k = 1}^{n} k^{4} + 10 \sum_{k = 1}^{n} k^{3} + 10 \sum_{k = 1}^{n} k^{2} + 5 \sum_{k = 1}^{n} k + \sum_{k = 1}^{n} 1 \Rightarrow n^{5} + 5 n^{4} + 10 n^{3} + 10 n^{2} + 5 n = 5 \sum_{k = 1}^{n} k^{4} + \frac{10 n^{2} {(n + 1)}^{2}}{4} + \frac{10 n (n + 1) (2 n + 1)}{6} + \frac{5 n (n + 1)}{2} + n \Rightarrow 5 \sum_{k = 1}^{n} k^{4} = n^{5} + 5 n^{4} + 10 n^{3} + 10 n^{2} + 4 n - \frac{5 n^{2} {(n + 1)}^{2}}{2} - \frac{5 n (n + 1) (2 n + 1)}{3} - \frac{5 n (n + 1)}{2} \Rightarrow 5 \sum_{k = 1}^{n} k^{4} = n^{5} + \frac{5 n^{4}}{2} + \frac{5 n^{3}}{3} - \frac{n}{6}$

This expression on further simplification gives

$\sum_{k = 1}^{n} k^{4} = \frac{n (n + 1) (2 n + 1) (3 n^{2} + 3 n - 1)}{30}$

$∴ \lim_{n \to \infty} \frac{1^{4} + 2^{4} + 3^{4} + . . . + n^{4}}{n^{5}} - \lim_{n \to \infty} \frac{1^{3} + 2^{3} + . . . + n^{3}}{n^{5}} = \lim_{n \to \infty} \frac{n (n + 1) (2 n + 1) (3 n^{2} + 3 n - 1)}{30 n^{5}} - \lim_{n \to \infty} \frac{n^{2} {(n + 1)}^{2}}{4 n^{5}} = \frac{1}{30} \lim_{n \to \infty} (1 + \frac{1}{n}) (2 + \frac{1}{n}) (3 + \frac{3}{n} - \frac{1}{n^{2}}) - \frac{1}{4} \lim_{n \to \infty} \frac{1}{n} {(1 + \frac{1}{n})}^{2} = \frac{1}{30} \times (1 + 0) \times (2 + 0) \times (3 + 0 - 0) - \frac{1}{4} \times 0 (\lim_{n \to \infty} \frac{1}{n} = \lim_{n \to \infty} \frac{1}{n^{2}} = . . . = 0)$
$= \frac{1}{30} \times 6 - 0 = \frac{1}{5}$

Page No 29.39:

Question 26:

Evaluate: $\lim_{n \to \infty} \frac{1.2 + 2.3 + 3.4 + . . . + n (n + 1)}{n^{3}}$

Answer:

$\lim_{n \to \infty} \frac{1.2 + 2.3 + 3.4 + . . . + n (n + 1)}{n^{3}} = \lim_{n \to \infty} \frac{\sum_{k = 1}^{n} k (k + 1)}{n^{3}} = \lim_{n \to \infty} \frac{\sum_{k = 1}^{n} k^{2} + \sum_{k = 1}^{n} k}{n^{3}} = \lim_{n \to \infty} \frac{\frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}}{n^{3}}$
$= \lim_{n \to \infty} \frac{n (n + 1) (2 n + 1 + 3)}{6 n^{3}} = \lim_{n \to \infty} \frac{2 n (n + 1) (n + 2)}{6 n^{3}} = \frac{1}{3} \lim_{n \to \infty} (1 + \frac{1}{n}) (1 + \frac{2}{n}) = \frac{1}{3} \times (1 + 0) \times (1 + 0) (\lim_{n \to \infty} \frac{1}{n} = 0) = \frac{1}{3}$

Page No 29.49:

Question 1:

$\lim_{x \to 0} \frac{\sin 3 x}{5 x}$

Answer:

$\lim_{x \to 0} [\frac{\sin 3 x}{5 x}]$

= $\frac{1}{5} \lim_{x \to 0} [\frac{\sin 3 x}{3 x} \times 3]$ $[∵ \lim_{x \to 0} (\frac{\sin 3 x}{3 x}) = 1]$

= $\frac{1}{5} \times 1 \times 3$

= $\frac{3}{5}$

Page No 29.49:

Question 2:

$\lim_{x \to 0} \frac{\sin x^{0}}{x}$

Answer:

We know that $x ° = \frac{π}{180} x$ .

$∴ \lim_{x \to 0} \frac{\sin x^{0}}{x} = \lim_{x \to 0} \frac{\sin \frac{π}{180} x}{x} = \lim_{x \to 0} \frac{\sin (\frac{π}{180} x)}{(\frac{π}{180} x)} \times \frac{π}{180} = \frac{π}{180} \times 1 = \frac{π}{180}$

Page No 29.49:

Question 3:

$\lim_{x \to 0} \frac{x^{2}}{\sin x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{x^{2}}{\sin x^{2}}]$

$When x \to 0, then x^{2} \to 0 .$

Let $θ = x^{2}$

⇒ $\lim_{θ \to 0} (\frac{θ}{\sin θ})$

= 1

Page No 29.49:

Question 4:

$\lim_{x \to 0} \frac{\sin x \cos x}{3 x}$

Answer:

$\lim_{x \to 0} (\frac{\sin x \cos x}{3 x})$

= $\frac{1}{3} \lim_{x \to 0} (\frac{\sin x}{x}) \times \cos x$

= $\frac{1}{3} \times 1 \times \cos 0$

= $\frac{1}{3} \times 1$

= $\frac{1}{3}$

Page No 29.50:

Question 5:

$\lim_{x \to 0} \frac{3 \sin x - 4 \sin^{3} x}{x}$

Answer:

$\lim_{x \to 0} [\frac{3 \sin x - 4 \sin^{3} x}{x}]$

= $\lim_{x \to 0} [\frac{\sin 3 x}{x}]$ $[∵ \sin^{3} A = 3 sinA - 4 \sin^{3} A]$

= $\lim_{x \to 0} [\frac{\sin 3 x}{3 x} \times 3]$

= 1 × 3

= 3

Page No 29.50:

Question 6:

$\lim_{x \to 0} \frac{\tan 8 x}{\sin 2 x}$

Answer:

$\lim_{x \to 0} [\frac{\tan 8 x}{\sin 2 x}]$

$\Rightarrow \lim_{x \to 0} [\frac{\tan 8 x}{8 x} \times \frac{8 x}{\frac{\sin 2 x}{2 x} \times 2 x}] [∵ \lim_{x \to 0} \frac{\tan 8 x}{8 x} = 1, \lim_{x \to 0} \frac{\sin 2 x}{2 x} = 1] \Rightarrow 4$

Page No 29.50:

Question 7:

$\lim_{x \to 0} \frac{\tan m x}{\tan n x}$

Answer:

$\lim_{x \to 0} [\frac{\tan m x}{\tan n x}]$

$\Rightarrow \lim_{x \to 0} (\frac{\tan m x}{m x} \times \frac{m x}{n x} \times \frac{n x}{\tan n x}) \Rightarrow \lim_{x \to 0} (\frac{\tan m x}{m x} \times \frac{m x}{\frac{\tan n x}{n x}} \times n x) \Rightarrow \frac{m}{n} [∵ \lim_{x \to 0} \frac{\tan x}{x} = 1]$

Page No 29.50:

Question 8:

$\lim_{x \to 0} \frac{\sin 5 x}{\tan 3 x}$

Answer:

$\lim_{x \to 0} [\frac{\sin 5 x}{\tan 3 x}]$

$\Rightarrow \lim_{x \to 0} [\lim_{x \to 0} \frac{\sin 5 x}{5 x} \times \frac{5 x}{\frac{\tan 3 x}{3 x} \times 3 x}] \Rightarrow \frac{5}{3} [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1, \lim_{x \to 0} \frac{\tan x}{x} = 1]$

Page No 29.50:

Question 9:

$\lim_{x \to 0} \frac{\sin x^{n}}{x^{n}}$

Answer:

$\lim_{x \to 0} \frac{\sin x^{n}}{x^{n}}$

It is of the form $(\frac{0}{0})$ .

$Let y = x^{n} \Rightarrow \lim_{x \to 0} = \lim_{y \to 0} \Rightarrow \lim_{y \to 0} \frac{\sin y}{y} \Rightarrow 1 \{∵ \lim_{x \to 0} \frac{\sin x}{x} = 1\}$

Page No 29.50:

Question 10:

$\lim_{x \to 0} \frac{7 x \cos x - 3 \sin x}{4 x + \tan x}$

Answer:

$\lim_{x \to 0} [\frac{7 x \cos x - 3 \sin x}{4 x + \tan x}]$

It is of the form $(\frac{0}{0})$ .
Dividing the numerator and the denominator by x:

$\Rightarrow \lim_{x \to 0} \frac{7 \cos x - 3 (\frac{\sin x}{x})}{4 + (\frac{\tan x}{x})} \Rightarrow \frac{7 \lim_{x \to 0} (\cos x) - 3 \lim_{x \to 0} (\frac{\sin x}{x})}{4 + \lim_{x \to 0} (\frac{\tan x}{x})} \Rightarrow \frac{7.1 - 3.1}{4 + 1} \Rightarrow \frac{4}{5}$

Page No 29.50:

Question 11:

$\lim_{x \to 0} \frac{\cos a x - \cos b x}{\cos c x - \cos d x}$

Answer:

$\lim_{x \to 0} [\frac{\cos a x - \cos b x}{\cos c x - \cos d x}]$

It is of the form $(\frac{0}{0})$ .

$\Rightarrow \lim_{x \to 0} [\frac{- 2 \sin (\frac{a x + b x}{2}) \sin (\frac{a x - b x}{2})}{- 2 \sin (\frac{c x + d x}{a 2}) \sin (\frac{c x - d x}{2})}] = \lim_{x \to 0} [\frac{\frac{\sin (\frac{a x + b x}{2})}{(\frac{a x + b}{x})} \times (\frac{a x + b x}{2}) \times \frac{\sin (\frac{a x - b x}{2})}{(\frac{a x - b x}{2})} \times (\frac{a x - b x}{2})}{\frac{\sin (\frac{c x + d x}{2})}{(\frac{c x + d x}{2})} \times (\frac{c x + d x}{2}) \times \frac{\sin (\frac{c x - d x}{2})}{(\frac{c x - d x}{2})} \times (\frac{c x - d x}{2})}] = 1 \times \lim_{x \to 0} [\frac{(\frac{a x + b x}{2}) (\frac{a x - b x}{2})}{(\frac{c x + d x}{2}) (\frac{c x - d x}{2})}] = \lim_{x \to 0} \frac{x^{2} (a + b) (a - b)}{x^{2} (c + d) (c - d)} = \frac{a^{2} - b^{2}}{c^{2} - d^{2}}$

Page No 29.50:

Question 12:

$\lim_{x \to 0} \frac{\tan^{2} 3 x}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{\tan^{2} 3 x}{x^{2}}]$

$= \lim_{x \to 0} [\frac{\tan 3 x}{3 x} \times \frac{\tan 3 x}{3 x}] \times 9 = 1 \times 1 \times 9 [∵ \lim_{x \to 0} \frac{\tan x}{x} = 1] = 9$

Page No 29.50:

Question 13:

$\lim_{x \to 0} \frac{1 - \cos m x}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{1 - \cos (m x)}{x^{2}}]$

$= \lim_{x \to 0} [\frac{2 \sin^{2} (\frac{m x}{2})}{x^{2}}] [∵ 1 - \cos A = 2 \sin^{2} (\frac{A}{2})] = 2 \lim_{x \to 0} [\frac{\sin (\frac{m x}{2})}{\frac{m x}{2}} \times \frac{\sin (\frac{m x}{2})}{\frac{m x}{2}}] \times \frac{m}{2} \times \frac{m}{2} = 2 \times \frac{m}{2} \times \frac{m}{2} [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] = \frac{m^{2}}{2}$

Page No 29.50:

Question 14:

$\lim_{x \to 0} \frac{3 \sin 2 x + 2 x}{3 x + 2 \tan 3 x}$

Answer:

$\lim_{x \to 0} [\frac{3 \sin (2 x) + 2 x}{3 x + 2 \tan (3 x)}]$

$= \lim_{x \to 0} [\frac{\frac{3 \sin 2 x}{x} + 2}{3 + \frac{2 \tan 3 x}{x}}] = \frac{3 \lim_{x \to 0} (\frac{\sin 2 x}{2 x}) \times 2 + 2}{3 + 2 \lim_{x \to 0} (\frac{\tan 3 x}{x}) \times 3} = \frac{6 + 2}{3 + 6} = \frac{8}{9}$

Page No 29.50:

Question 15:

$\lim_{x \to 0} \frac{\cos 3 x - \cos 7 x}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{\cos 3 x - \cos 7 x}{x^{2}}]$
$= \lim_{x \to 0} [\frac{- 2 \sin (\frac{3 x + 7 x}{2}) \sin \frac{(3 x - 7 x)}{2}}{x^{2}}] [\cos C - \cos D = - 2 \sin (\frac{C + D}{2}) \sin (\frac{C - D}{2})] = \lim_{x \to 0} [\frac{- 2 \sin 5 x \sin (- 2 x)}{x^{2}}] = \lim_{x \to 0} [\frac{2 \sin 5 x \sin 2 x}{x^{2}}] [∵ \sin (- θ) = - \sin θ] = 2 \lim_{x \to 0} [\frac{\sin 5 x}{5 x} \times \frac{\sin 2 x}{2 x}] \times 5 \times 2 = 2 \times 5 \times 2 = 20$

Page No 29.50:

Question 16:

$\lim_{θ \to 0} \frac{\sin 3 θ}{\tan 2 θ}$

Answer:

$\lim_{θ \to 0} [\frac{\sin 3 θ}{\tan 2 θ}]$

$= \lim_{θ \to 0} [\frac{\sin 3 θ}{3 θ} \times \frac{3 θ}{\frac{\tan 2 θ}{2 θ} \times 2 θ}] = \frac{3}{2}$

Page No 29.50:

Question 17:

$\lim_{x \to 0} \frac{\sin x^{2} (1 - \cos x^{2})}{x^{6}}$

Answer:

$\lim_{x \to 0} [\frac{\sin x^{2} (1 - \cos x^{2})}{x^{6}}]$
$= \lim_{x \to 0} [\frac{\sin x^{2} \times 2 \sin^{2} (\frac{x^{2}}{2})}{x^{6}}] [∵ 1 - \cos A = 2 \sin^{2} (\frac{A}{2})] = 2 \lim_{x \to 0} [\frac{\sin x^{2}}{x^{2}} \times \frac{\sin (\frac{x^{2}}{2})}{2 \times \frac{x^{2}}{2}} \times \frac{\sin (\frac{x^{2}}{2})}{2 \times \frac{x^{2}}{2}}] = \frac{2}{2 \times 2} = \frac{1}{2}$

Page No 29.50:

Question 18:

$\lim_{x \to 0} \frac{\sin^{2} 4 x^{2}}{x^{4}}$

Answer:

$\lim_{x \to 0} [\frac{\sin^{2} 4 x^{2}}{x^{4}}]$

$= \lim_{x \to 0} [\frac{\sin (4 x^{2})}{x^{2}} \times \frac{\sin (4 x^{2})}{x^{2}}] = \lim_{x \to 0} [\frac{\sin (4 x^{2})}{4 x^{2}} \times 4 \times \frac{\sin (4 x^{2})}{4 x^{2}} \times 4] = 4 \times 4 [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] = 16$

Page No 29.50:

Question 19:

$\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^{2} + \tan x}$

Answer:

$\lim_{x \to 0} [\frac{x \cos x + 2 \sin x}{x^{2} + \tan x}]$

Dividing the numerator and the denominator by x, we get:

$\lim_{x \to 0} [\frac{\cos x + \frac{2 \sin x}{x}}{x + \frac{\tan x}{x}}] = \frac{\cos 0 + 2 \times 1}{0 + 1} [∵ \lim_{x \to 0} (\frac{\sin x}{x}) = 1, \lim_{x \to 0} (\frac{\tan x}{x}) = 1] = \frac{3}{1}$

Page No 29.50:

Question 20:

$\lim_{x \to 0} \frac{2 x - \sin x}{\tan x + x}$

Answer:

$\lim_{x \to 0} [\frac{2 x - \sin x}{\tan x + x}]$

Dividing the numerator and the denominator by x, we get:

$\lim_{x \to 0} [\frac{2 - \frac{\sin x}{x}}{\frac{\tan x}{x} + 1}] = \frac{2 - 1}{1 + 1} [∵ \lim_{x \to 0} \frac{\tan x}{x} = 1, \lim_{x \to 0} \frac{\sin x}{x} = 1] = \frac{1}{2}$

Page No 29.50:

Question 21:

$\lim_{x \to 0} \frac{5 x \cos x + 3 \sin x}{3 x^{2} + \tan x}$

Answer:

$\lim_{x \to 0} [\frac{5 x \cos x + 3 \sin x}{3 x^{2} + \tan x}]$

Dividing the numerator and the denominator by x:

$\lim_{x \to 0} [\frac{5 \cos x + 3 (\frac{\sin x}{x})}{3 x + (\frac{\tan x}{x})}] = [\frac{5 \cos 0 + 3}{3 \times 0 + 1}] [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1, \lim_{x \to 0} \frac{\tan x}{x} = 1] = \frac{5 + 3}{0 + 1} = 8$

Page No 29.50:

Question 22:

$\lim_{x \to 0} \frac{\sin 3 x - \sin x}{\sin x}$

Answer:

$\lim_{x \to 0} [\frac{\sin 3 x - \sin x}{\sin x}]$

$= \lim_{x \to 0} [\frac{2 \cos (\frac{3 x + x}{2}) \sin (\frac{3 x - x}{2})}{\sin x}] [∵ \sin C - \sin D = 2 \cos (\frac{C + D}{2}) \sin (\frac{C - D}{2})] = \lim_{x \to 0} [\frac{2 \cos 2 x . \sin x}{\sin x}] = 2 \cos 0 = 2$

Page No 29.50:

Question 23:

$\lim_{x \to 0} \frac{\sin 5 x - \sin 3 x}{\sin x}$

Answer:

$\lim_{x \to 0} [\frac{\sin 5 x - \sin 3 x}{\sin x}]$

$= \lim_{x \to 0} [\frac{2 \cos (\frac{5 x + 3 x}{2}) \sin (\frac{5 x - 3 x}{2})}{\sin x}] [∵ sinC - sinD = 2 \cos (\frac{C + D}{2}) \sin (\frac{C - D}{2})] = \lim_{x \to 0} [\frac{2 \cos 4 x . \sin x}{\sin x}] = 2 \cos 0 = 2$

Page No 29.50:

Question 24:

$\lim_{x \to 0} \frac{\cos 3 x - \cos 5 x}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{\cos 3 x - \cos 5 x}{x^{2}}]$

$= \lim_{x \to 0} [\frac{- 2 \sin (\frac{3 x + 5 x}{2}) \sin (\frac{3 x - 5 x}{2})}{x^{2}}] [∵ cosC - cosD = - 2 \sin (\frac{C + D}{2}) \sin (\frac{C - D}{2})] = 2 \lim_{x \to 0} [\frac{2 \sin 4 x \sin x}{x^{2}}] [∵ \sin (- θ) = - \sin θ] = 2 \lim_{x \to 0} [\frac{\sin 4 x}{4 x} \times 4 \times \frac{\sin x}{x}] = 2 \times 4 = 8 .$

Page No 29.50:

Question 25:

$\lim_{x \to 0} \frac{\tan 3 x - 2 x}{3 x - \sin^{2} x}$

Answer:

$\lim_{x \to 0} [\frac{\tan 3 x - 2 x}{3 x - \sin^{2} x}]$

Dividing the numerator and the denominator by x:

$\lim_{x \to 0} [\frac{\frac{\tan 3 x}{x} - 2}{3 - \frac{\sin^{2} x}{x}}] = \lim_{x \to 0} [\frac{\frac{\tan 3 x}{3 x} \times 3 - 2}{3 - (\frac{\sin x}{x}) \times \sin x}] = \frac{3 - 2}{3 - 1 \times 0} [∵ \lim_{x \to 0} \frac{\tan x}{x} = 1, \lim_{x \to 0} \frac{\sin x}{x} = 1] = \frac{1}{3}$

Page No 29.50:

Question 26:

$\lim_{x \to 0} \frac{\sin (2 + x) - \sin (2 - x)}{x}$

Answer:

$\lim_{x \to 0} [\frac{\sin (2 + x) - \sin (2 - x)}{x}]$

$\Rightarrow \lim_{x \to 0} [\frac{2 \cos (\frac{2 + x + 2 - x}{2}) \sin (\frac{2 + x - 2 + x}{2})}{x}] \Rightarrow \lim_{x \to 0} [\frac{2 \cos 2 \sin x}{x}] \Rightarrow 2 \cos 2 [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1]$

Page No 29.50:

Question 27:

$\lim_{h \to 0} \frac{{(a + h)}^{2} \sin (a + h) - a^{2} \sin a}{h}$

Answer:

$\lim_{h \to 0} [\frac{{(a + h)}^{2} \sin (a + h) - a^{2} \sin a}{h}]$

$= \lim_{h \to 0} [\frac{a^{2} \sin (a + h) + h^{2} \sin (a + h) + 2 a h \sin (a + h) - a^{2} \sin a}{h}] = \lim_{h \to 0} [a^{2} \{\frac{\sin (a + h) - \sin (a)}{h}\} + \frac{h^{2} \sin (a + h)}{h} + \frac{2 a h \sin (a + h)}{h}] Div iding and multiplying the denominator by 2: \lim_{h \to 0} [a^{2} \{\frac{2 \cos (\frac{a + h + a}{2}) \sin (\frac{a + h - a}{2})}{2 \times \frac{h}{2}}\} + h \sin (a + h) + 2 a \sin (a + h)] = \lim_{h \to 0} [a^{2} \cos (\frac{a + h + a}{2}) + h \sin (a + h) + 2 a \sin (a + h)] = a^{2} \cos a + 0 + 2 a \sin a = a^{2} \cos a + 2 a \sin a$

Page No 29.50:

Question 28:

$\lim_{x \to 0} \frac{\tan x - \sin x}{\sin 3 x - 3 \sin x}$

Answer:

$\lim_{x \to 0} [\frac{\tan x - \sin x}{\sin 3 x - 3 \sin x}]$

$= \lim_{x \to 0} [\frac{\frac{si n x}{\cos x} - \sin x}{3 \sin x - 4 \sin^{3} x - 3 \sin x}] = \lim_{x \to 0} [\frac{\sin x (1 - \cos x)}{\cos x \times - 4 \sin^{3} x}] = \lim_{x \to 0} [\frac{2 \sin^{2} (\frac{x}{2})}{\cos x \times - 4 \sin^{2} x}] = \lim_{x \to 0} [\frac{2 \sin (\frac{x}{2}) \times \sin (\frac{x}{2})}{\cos x \times (- 4 \sin x) \times \sin x}] = \lim_{x \to 0} [\frac{2 \frac{\sin (\frac{x}{2})}{\frac{x}{2}} \times \frac{x}{2} \times \frac{\sin (\frac{x}{2})}{\frac{x}{2}} \times \frac{x}{2}}{\cos x \times - 4 \frac{\sin x}{x} \times x \times \frac{\sin x}{x} \times x}] = \lim_{x \to 0} [\frac{2 \times \frac{x}{2} \times \frac{x}{2}}{\cos x \times - 4 \times x \times x}] = \frac{- 1}{8 \cos 0} = - \frac{1}{8}$

Page No 29.50:

Question 29:

$\lim_{x \to 0} \frac{\sec 5 x - \sec 3 x}{\sec 3 x - \sec x}$

Answer:

$\lim_{x \to 0} [\frac{\sec 5 x - \sec 3 x}{\sec 3 x - \sec x}]$

$= \lim_{x \to 0} [\frac{\frac{1}{\cos 5 x} \frac{1}{\cos 3 x}}{\frac{1}{\cos 3 x} - \frac{1}{\cos x}}] = \lim_{x \to 0} [\frac{\cos 3 x - \cos 5 x}{\cos 5 x \cos 3 x \{\frac{\cos x - \cos 3 x}{\cos x \cos 3 x}\}}] = \lim_{x \to 0} [\frac{(\cos 3 x - \cos 5 x) \cos x}{\cos 5 x \{\cos x - \cos 3 x\}}] = \lim_{x \to 0} [\frac{- 2 \sin (\frac{3 x + 5 x}{2}) \sin (\frac{3 x - 5 x}{2}) \times \cos x}{\cos (5 x) [- 2 \sin (\frac{x + 3 x}{2}) \sin (\frac{x - 3 x}{2})]}] [∵ cosC - cosD = - 2 \sin (\frac{C + D}{2}) \sin (\frac{C - D}{2})] = \lim_{x \to 0} [\frac{\sin (4 x) \times \sin (- x) \times \cos x}{\cos (5 x) \times \sin (2 x) \times \sin (- x)}] = \lim_{x \to 0} [\frac{\sin (4 x)}{4 x} \times \frac{4 x}{\frac{\sin (2 x)}{2 x} \times 2 x} \times \frac{\cos x}{\cos 5 x}] = \frac{4}{2} \frac{\cos 0}{\cos 0} = 2$

Page No 29.50:

Question 30:

$\lim_{x \to 0} \frac{1 - \cos 2 x}{\cos 2 x - \cos 8 x}$

Answer:

$\lim_{x \to 0} [\frac{1 - \cos 2 x}{\cos 2 x - \cos 8 x}]$

$= \lim_{x \to 0} [\frac{2 \sin^{2} x}{2 \sin (\frac{2 x + 8 x}{2}) \sin (\frac{8 x - 2 x}{2})}] [∵ cosC - cosD = 2 \sin (\frac{C + D}{2}) \sin (\frac{D - C}{2})] = \lim_{x \to 0} [\frac{2 \sin x \times \sin x}{2 \sin 5 x \times \sin 3 x}] = \lim_{x \to 0} [\frac{\sin x \times \sin x}{\sin 5 x \times \sin 3 x}] [∵ \sin (- θ) = - \sin θ] = \lim_{x \to 0} [\frac{\frac{\sin x}{x} \times x \times \frac{\sin x}{x} \times x}{\frac{\sin 5 x}{5 x} \times 5 x \times \frac{\sin 3 x \times 3 x}{3 x}}] = \frac{1}{15}$

Page No 29.50:

Question 31:

$\lim_{x \to 0} \frac{1 - \cos 2 x + \tan^{2} x}{x \sin x}$

Answer:

$\lim_{x \to 0} [\frac{1 - \cos 2 x + \tan^{2} x}{x \sin x}] = \lim_{x \to 0} [\frac{2 \sin^{2} x + \tan^{2} x}{x \sin x}] [∵ 1 - \cos 2 A = 2 \sin^{2} A] Dividing numerator & denominator by x^{2} : \lim_{x \to 0} [\frac{\frac{2 \sin^{2} x}{x^{2}} + \frac{\tan^{2} x}{x^{2}}}{(\frac{\sin x}{x})}] = \frac{2 {(1)}^{2} + {(1)}^{2}}{1} [∵ \lim_{x \to 0} \frac{\sin^{2} x}{x^{2}} = 1, \lim_{x \to 0} \frac{\tan^{2} x}{x^{2}} = 1] = 3$

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Question 32:

$\lim_{x \to 0} \frac{\sin (a + x) + \sin (a - x) - 2 \sin a}{x \sin x}$

Answer:

$\lim_{x \to 0} [\frac{\sin (a + x) + \sin (a - x) - 2 \sin a}{x \sin x}] = \lim_{x \to 0} [\frac{2 \sin (\frac{a + x + a - x}{2}) \cos (\frac{a + x - a + x}{2}) - 2 \sin a}{x \sin x}] \{∵ \sin C + \sin D = 2 \sin (\frac{C + D}{2}) \cos (\frac{C - D}{2})\} = \lim_{x \to 0} [\frac{2 \sin a \cos x - 2 \sin a}{x \sin x}] = \lim_{x \to 0} [\frac{2 \sin a (\cos x - 1)}{x \sin x}] = \lim_{x \to 0} [\frac{2 \sin a (1 - 2 \sin^{2} \frac{x}{2} - 1)}{x \sin x}] = \lim_{x \to 0} [\frac{2 \sin a (- 2 \sin^{2} \frac{x}{2})}{x \sin x}] = - 4 \sin a \lim_{x \to 0} [\frac{1}{\frac{x \sin x}{x^{2}}} \times \frac{\sin (\frac{x}{2})}{\frac{x}{2}} \times \frac{\sin (\frac{x}{2})}{\frac{x}{2}} \times \frac{1}{4}] = - 4 \sin a \times \frac{1}{1} \times 1 \times 1 \times \frac{1}{4} = - \sin a$

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Question 33:

$\lim_{x \to 0} \frac{x^{2} - \tan 2 x}{\tan x}$

Answer:

$\lim_{x \to 0} [\frac{x^{2} - \tan 2 x}{\tan x}] Dividing the numerator and the denominator by x: \lim_{x \to 0} [\frac{x - \frac{\tan 2 x}{x}}{(\frac{\tan x}{x})}] = \lim_{x \to 0} [\frac{x - \frac{\tan 2 x}{2 x} \times 2}{(\frac{\tan x}{x})}] = \frac{0 - 2 (1)}{1} = - 2$

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Question 34:

$\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{2} - \sqrt{1 + \cos x}}{x^{2}}] Rationalising the numerator : \lim_{x \to 0} [\frac{(\sqrt{2} - \sqrt{1 + \cos x}) (\sqrt{2} + \sqrt{1 + \cos x})}{x^{2} \times (\sqrt{2} + \sqrt{1 + \cos x})}] = \lim_{x \to 0} [\frac{2 - (1 + \cos x)}{x^{2} [\sqrt{2} + \sqrt{1 + \cos x}]}] = \lim_{x \to 0} [\frac{1 - \cos x}{x^{2} [\sqrt{2} + \sqrt{1 + \cos x}]}] = \lim_{x \to 0} [\frac{2 \sin^{2} (\frac{x}{2})}{x^{2} \{\sqrt{2} + \sqrt{1 + \cos x}\}}] = 2 \lim_{x \to 0} [\frac{\sin^{2} (\frac{x}{2})}{4 \times {(\frac{x}{2})}^{2}} \times \frac{1}{\sqrt{2} + \sqrt{1 + \cos x}}] = 2 \times \frac{1}{4} \times \frac{1}{\sqrt{2} + \sqrt{1 + \cos 0}} = \frac{2}{4 (\sqrt{2} + \sqrt{2})} = \frac{1}{4 \sqrt{2}}$

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Question 35:

$\lim_{x \to 0} \frac{x \tan x}{1 - \cos x}$

Answer:

$\lim_{x \to 0} [\frac{x \tan x}{1 - \cos x}] {Dividing the numerator and the denominator by x}^{2} : \lim_{x \to 0} [\frac{\frac{x \tan x}{x^{2}}}{\frac{1 - \cos x}{x^{2}}}] = \lim_{x \to 0} [\frac{\frac{\tan x}{x}}{\frac{2 \sin^{2} \frac{x}{2}}{\frac{x}{2} \times \frac{x}{2} \times 4}}] = \frac{4}{2} = 2$

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Question 36:

$\lim_{x \to 0} \frac{x^{2} + 1 - \cos x}{x \sin x}$

Answer:

$\lim_{x \to 0} [\frac{x^{2} + 1 - \cos x}{x \sin x}] = \lim_{x \to 0} [\frac{x^{2} + 2 \sin^{2} \frac{x}{2}}{x \sin x}] Dividing the numerator and the denominator by x^{2} : \lim_{x \to 0} [\frac{1 + \frac{2 \sin^{2} (\frac{x}{2})}{x^{2}}}{\frac{\sin x}{x}}] = \lim_{x \to 0} [\frac{1 + \frac{2 \sin^{2} (\frac{x}{2})}{{(\frac{x}{2})}^{2} \times 4}}{\frac{\sin x}{x}}] = \frac{1 + \frac{2}{4}}{1} = \frac{\frac{3}{2}}{1} = \frac{3}{2}$

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Question 37:

$\lim_{x \to 0} \frac{\sin 2 x (\cos 3 x - \cos x)}{x^{3}}$

Answer:

$\lim_{x \to 0} [\frac{\sin 2 x (\cos 3 x - \cos x)}{x^{3}}] = \lim_{x \to 0} [\frac{\sin 2 x \times - 2 \sin (\frac{3 x + x}{2}) \sin (\frac{3 x - x}{2})}{x^{3}}] = - 2 \lim_{x \to 0} [\frac{\sin 2 x}{2 x} \times \frac{\sin 2 x}{2 x} \times \frac{\sin x}{x}] \times 2 \times 2 = - 8$

Page No 29.50:

Question 38:

$\lim_{x \to 0} \frac{2 \sin x^{\circ} - \sin 2 x^{\circ}}{x^{3}}$

Answer:

$\lim_{x \to 0} [\frac{2 \sin x ° - \sin (2 x °)}{x^{3}}] = \lim_{x \to 0} [\frac{2 \sin (\frac{π x}{180}) - \sin (\frac{2 π x}{180})}{x^{3}}] = \lim_{x \to 0} [\frac{2 \sin (\frac{π x}{180}) - 2 \sin (\frac{π x}{180}) \times \cos (\frac{π x}{180})}{x^{3}}] = \lim_{x \to 0} [\frac{2 \sin (\frac{π x}{180}) [1 - \cos (\frac{π x}{180})]}{x^{3}}] = \lim_{x \to 0} [\frac{2 \sin (\frac{π x}{180}) \times 2 \sin^{2} (\frac{π x}{360})}{x \times x^{2}}] = \lim_{x \to 0} [\frac{4 \sin [\frac{π x}{180}] \times \sin^{2} [\frac{π x}{360}]}{\frac{π x}{180} \times \frac{π x}{360} \times \frac{π x}{360}} \times \frac{π}{180} \times {(\frac{π}{360})}^{2}] = 4 \lim_{x \to 0} [\frac{\sin (\frac{π x}{180})}{\frac{π x}{180}} \times \frac{\sin (\frac{π x}{360}) \times \sin (\frac{π x}{360})}{\frac{π x}{360} \times \frac{π x}{360}} \times \frac{π^{3}}{180 \times 360^{2}}] = 4 \times 1 \times 1 \times 1 \times \frac{π^{3}}{180 \times 360 \times 360} = {(\frac{π}{180})}^{3}$

Page No 29.50:

Question 39:

$\lim_{x \to 0} \frac{x^{3} \cot x}{1 - \cos x}$

Answer:

$\lim_{x \to 0} [\frac{x^{3} \cot x}{1 - \cos x}] = \lim_{x \to 0} [\frac{x^{3}}{\tan x (1 - \cos x)}] = \lim_{x \to 0} [\frac{x^{3}}{\tan x \times 2 \sin^{2} (\frac{x}{2})}] = \lim_{x \to 0} [\frac{x}{\tan x} \times \frac{x^{2}}{2 \sin^{2} (\frac{x}{2})}] = \lim_{x \to 0} [\frac{x}{\tan x} \times \frac{\frac{x^{2}}{4} \times 4}{2 \times \sin^{2} \frac{x}{2}}] = \lim_{x \to 0} [\frac{x}{\tan x} \times {(\frac{\frac{x}{2}}{\sin \frac{x}{2}})}^{2} \times \frac{4}{2}] = 1 \times 1 \times \frac{4}{2} = 2$

Page No 29.50:

Question 40:

$\lim_{x \to 0} \frac{x \tan x}{1 - \cos 2 x}$

Answer:

$\lim_{x \to 0} [\frac{x \tan x}{1 - \cos (2 x)}] = \lim_{x \to 0} [\frac{x \tan x}{2 \sin^{2} x}] Dividing the numerator and the denominator by x^{2} : \lim_{x \to 0} [\frac{\frac{x \tan x}{x^{2}}}{2 \frac{\sin^{2} x}{x^{2}}}] = \lim_{x \to 0} [\frac{\frac{\tan x}{x}}{2 {(\frac{\sin x}{x})}^{2}}] = \frac{1}{2}$

Page No 29.50:

Question 41:

$\lim_{x \to 0} \frac{\sin (3 + x) - \sin (3 - x)}{x}$

Answer:

$\lim_{x \to 0} [\frac{\sin (3 + x) - \sin (3 - x)}{x}] = \lim_{x \to 0} [\frac{2 \cos (\frac{3 + x + 3 - x}{2}) \sin (\frac{3 + x - 3 + x}{2})}{x}] \{∵ \sin C - \sin D = 2 \cos (\frac{C + D}{2}) \sin (\frac{C - D}{2})\} = \lim_{x \to 0} [\frac{2 \cos 3 \sin x}{x}] = 2 \cos 3$

Page No 29.50:

Question 42:

$\lim_{x \to 0} \frac{\cos 2 x - 1}{\cos x - 1}$

Answer:

$\lim_{x \to 0} [\frac{\cos (2 x) - 1}{\cos x - 1}] = \lim_{x \to 0} [\frac{1 - 2 \sin^{2} x - 1}{1 - 2 \sin^{2} (\frac{x}{2}) - 1}] = \lim_{x \to 0} [\frac{\sin^{2} x}{\sin^{2} (\frac{x}{2})}] = \lim_{x \to 0} [\frac{\sin^{2} x}{x^{2}} \times \frac{x^{2}}{\frac{\sin^{2} (\frac{x}{2})}{\frac{x^{2}}{4}} \times \frac{x^{2}}{4}}] = \lim_{x \to 0} [{(\frac{\sin x}{x})}^{2} \times \frac{1}{{(\frac{\sin (\frac{x}{2})}{\frac{x}{2}})}^{2}} \times 4] = 4$

Page No 29.51:

Question 43:

$\lim_{x \to 0} \frac{3 \sin^{2} x - 2 \sin x^{2}}{3 x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{3 \sin^{2} x - 2 \sin x^{2}}{3 x^{2}}] = \lim_{x \to 0} [\frac{3 \sin^{2} x}{3 x^{2}} - \frac{2 \sin x^{2}}{3 x^{2}}] = 1 - \frac{2}{3} = \frac{1}{3}$

Page No 29.51:

Question 44:

$\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}] = \lim_{x \to 0} [\frac{(\sqrt{1 + \sin x} - \sqrt{1 - \sin x}) (\sqrt{1 + \sin x} + \sqrt{1 - \sin x})}{x (\sqrt{1 + \sin x} + \sqrt{1 - \sin x})}] = \lim_{x \to 0} [\frac{(1 + \sin x) - (1 - \sin x)}{x (\sqrt{1 + \sin x} + \sqrt{1 - \sin x})}] = \lim_{x \to 0} [\frac{2 \sin x}{x (\sqrt{1 + \sin x} + \sqrt{1 - \sin x})}] = \frac{2}{\sqrt{1 + 0} + \sqrt{1 - 0}} = \frac{2}{2} = 1$

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Question 45:

$\lim_{x \to 0} \frac{1 - \cos 4 x}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{1 - \cos (4 x)}{x^{2}}] = \lim_{x \to 0} [\frac{2 \sin^{2} 2 x}{x^{2}}] = \lim_{x \to 0} 2 [\frac{\sin 2 x}{x} \times \frac{\sin 2 x}{x}] = \lim_{x \to 0} [2 \times \frac{\sin 2 x}{2 x} \times \frac{\sin 2 x}{2 x} \times 4] = 2 \times 1 \times 1 \times 4 = 8$

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Question 46:

$\lim_{x \to 0} \frac{x \cos x + \sin x}{x^{2} + \tan x}$

Answer:

$\lim_{x \to 0} [\frac{x \cos x + \sin x}{x^{2} + \tan x}] Dividing the numerator and the denominator by x: \lim_{x \to 0} [\frac{\cos x + \frac{\sin x}{x}}{x + \frac{\tan x}{x}}] = \frac{\cos 0 + 1}{0 + 1} = 2$

Page No 29.51:

Question 47:

$\lim_{x \to 0} \frac{1 - \cos 2 x}{3 \tan^{2} x}$

Answer:

$\lim_{x \to 0} [\frac{1 - \cos (2 x)}{3 \tan^{2} x}] = \lim_{x \to 0} [\frac{2 \sin^{2} x}{3 \tan^{2} x}] = \lim_{x \to 0} [\frac{2}{3} \times \frac{\sin^{2} x}{\sin^{2} x} \times \cos^{2} x] = \frac{2}{3} \cos^{2} 0 = \frac{2}{3}$

Page No 29.51:

Question 48:

$\lim_{θ \to 0} \frac{1 - \cos 4 θ}{1 - \cos 6 θ}$

Answer:

$\lim_{θ \to 0} [\frac{1 - \cos (4 θ)}{1 - \cos 6 θ}] = \lim_{θ \to 0} [\frac{2 \sin^{2} 2 θ}{2 \sin^{2} 3 θ}] \{∵ 1 - \cos A = 2 \sin^{2} (\frac{A}{2})\} = \lim_{θ \to 0} [\frac{\sin^{2} 2 θ}{{(2 θ)}^{2}} \times \frac{{(2 θ)}^{2}}{\frac{\sin^{2} 3 θ}{{(3 θ)}^{2}} \times {(3 θ)}^{2}}] = \lim_{θ \to 0} [{(\frac{\sin 2 θ}{2 θ})}^{2} \times {(\frac{3 θ}{\sin 3 θ})}^{2} \times \frac{4}{9}] = \frac{4}{9} [∵ \lim_{X \to 0} \frac{\sin x}{x} = 1]$

Page No 29.51:

Question 49:

$\lim_{x \to 0} \frac{a x + x \cos x}{b \sin x}$

Answer:

$\lim_{x \to 0} [\frac{a x + x \cos x}{b \sin x}] Dividing the numerator and the denominator by x: \lim_{x \to 0} [\frac{a + \cos x}{b (\frac{\sin x}{x})}] [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] = \frac{a + \cos 0}{b} = \frac{a + 1}{b}$

Page No 29.51:

Question 50:

$\lim_{θ \to 0} \frac{\sin 4 θ}{\tan 3 θ}$

Answer:

$\lim_{θ \to 0} [\frac{\sin 4 θ}{\tan 3 θ}] = \lim_{θ \to 0} [\frac{\sin 4 θ}{4 θ} \times \frac{4 θ}{\frac{\tan 3 θ}{3 θ} \times 3 θ}] = \frac{4}{3}$

Page No 29.51:

Question 51:

Evaluate the following limits:

$\lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}}$

Answer:

$\lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}} = \lim_{x \to 0} \frac{2 \sin x - 2 \sin x \cos x}{x^{3}} = \lim_{x \to 0} \frac{2 \sin x (1 - \cos x)}{x^{3}} = \lim_{x \to 0} \frac{2 \sin x \times 2 \sin^{2} \frac{x}{2}}{x^{3}}$
$= \lim_{x \to 0} \frac{\sin x \times \sin^{2} \frac{x}{2}}{x \times \frac{x^{2}}{4}} = \lim_{x \to 0} \frac{\sin x}{x} \times {(\lim_{x \to 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}})}^{2} = 1 \times 1 (\lim_{θ \to 0} \frac{\sin θ}{θ} = 1) = 1$

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Question 52:

$\lim_{x \to 0} \frac{1 - \cos 5 x}{1 - \cos 6 x}$

Answer:

$\lim_{x \to 0} [\frac{1 - \cos (5 x)}{1 - \cos (6 x)}] = \lim_{x \to 0} [\frac{2 \sin^{2} (\frac{5 x}{2})}{2 \sin^{2} 3 x}] = \lim_{x \to 0} [\frac{\sin^{2} (\frac{5 x}{2})}{{(\frac{5 x}{2})}^{2}} \times \frac{{(\frac{5 x}{2})}^{2}}{\frac{\sin^{2} 3 x}{{(3 x)}^{2}} \times {(3 x)}^{2}}] = \lim_{x \to 0} [{\{\frac{\sin (\frac{5 x}{2})}{\frac{5 x}{2}}\}}^{2} \frac{\frac{25}{4} x^{2}}{{\{\frac{\sin (3 x)}{3 x}\}}^{2} \times 9 x^{2}}] = \frac{25}{9 \times 4} = \frac{25}{36}$

Page No 29.51:

Question 53:

$\lim_{x \to 0} \frac{cosec x - \cot x}{x}$

Answer:

$\lim_{x \to 0} [\frac{cosec x - \cot x}{x}] = \lim_{x \to 0} [\frac{\frac{1}{\sin x} - \frac{\cos x}{\sin x}}{x}] = \lim_{x \to 0} [\frac{1 - \cos x}{x \sin x}] = \lim_{x \to 0} [\frac{2 \sin^{2} (\frac{x}{2})}{x \sin x}] = \lim_{x \to 0} [\frac{2 \sin^{2} (\frac{x}{2})}{{(\frac{x}{2})}^{2}} \times \frac{\frac{x^{2}}{4}}{\frac{x \sin x}{x \times x} \times x^{2}}] = 2 \times \frac{1}{4} = \frac{1}{2}$

Page No 29.51:

Question 54:

$\lim_{x \to 0} \frac{\sin 3 x + 7 x}{4 x + \sin 2 x}$

Answer:

$\lim_{x \to 0} [\frac{\sin 3 x + 7 x}{4 x + \sin 2 x}] = \lim_{x \to 0} [\frac{\frac{\sin 3 x}{3 x} \times 3 x + 7 x}{4 x + \frac{\sin 2 x}{2 x} \times 2 x}] = \lim_{x \to 0} [\frac{(\frac{\sin 3 x}{3 x} \times 3 + 7) x}{(4 + \frac{\sin 2 x}{2 x} \times 2) x}] = \frac{3 + 7}{4 + 2} = \frac{10}{6} = \frac{5}{3}$

Page No 29.51:

Question 55:

$\lim_{x \to 0} \frac{5 x + 4 \sin 3 x}{4 \sin 2 x + 7 x}$

Answer:

$\lim_{x \to 0} [\frac{5 x + 4 \sin 3 x}{4 \sin 2 x + 7 x}] = \lim_{x \to 0} [\frac{5 x + 4 \times \frac{\sin 3 x}{3 x} \times 3 x}{\frac{4 \sin 2 x}{2 x} \times 2 x + 7 x}] = \lim_{x \to 0} [\frac{(5 + 4 \frac{\sin 3 x \times 3}{3 x}) x}{(4 \frac{\sin 2 x}{2 x} \times 2 + 7) x}] = \frac{5 + 4 \times 3}{4 \times 2 + 7} [∵ \lim_{x \to 0} \frac{\sin (3 x)}{3 x} = 1] = \frac{17}{15}$

Page No 29.51:

Question 56:

$\lim_{x \to 0} \frac{3 \sin x - \sin 3 x}{x^{3}}$

Answer:

$\lim_{x \to 0} [\frac{3 \sin x - \sin (3 x)}{x^{3}}] = \lim_{x \to 0} [\frac{3 \sin x - (3 \sin x - 4 \sin^{3} x)}{x^{3}}] = \lim_{x \to 0} [\frac{4 \sin^{3} x}{x^{3}}] = \lim_{x \to 0} [4 {(\frac{\sin x}{x})}^{3}] = 4 \times 1 = 4$

Page No 29.51:

Question 57:

$\lim_{x \to 0} \frac{\tan 2 x - \sin 2 x}{x^{3}}$

Answer:

$\lim_{x \to 0} [\frac{\tan (2 x) - \sin (2 x)}{x^{3}}] = \lim_{x \to 0} [\frac{\frac{\sin 2 x}{\cos 2 x} - \sin 2 x}{x^{3}}] = \lim_{x \to 0} [\frac{\sin 2 x \{1 - \cos 2 x\}}{co S 2 x \times x^{3}}] = \lim_{x \to 0} [\frac{\sin 2 x \times 2 \sin^{2} x}{\cos 2 x \times x^{3}}] = \lim_{x \to 0} [\frac{\sin 2 x}{2 x} \times \frac{2}{\cos 2 x} \times 2 {(\frac{\sin x}{x})}^{2}] = \frac{2 \times 2}{\cos 0} = 4$

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Question 58:

$\lim_{x \to 0} \frac{\sin a x + b x}{a x + \sin b x}$

Answer:

$\lim_{x \to 0} [\frac{\sin (a x) + b x}{a x + \sin (b x)}] = \lim_{x \to 0} [\frac{\frac{\sin a x}{a x} \times a x + b x}{a x + \frac{\sin b x}{b x} \times b x}] = \lim_{x \to 0} [\frac{(\frac{\sin a x}{a x} \times a + b) x}{(a + \frac{\sin b x}{b x} \times b) x}] = \frac{1 \times a + b}{a + b} = 1$

Page No 29.51:

Question 59:

$\lim_{x \to 0} (cosec x - \cot x)$

Answer:

$\lim_{x \to 0} [cosec x - \cot x] = \lim_{x \to 0} [\frac{1}{\sin x} - \frac{\cos x}{\sin x}] = \lim_{x \to 0} [\frac{1 - \cos x}{\sin x}] [∵ 1 - \cos A = 2 \sin^{2} (\frac{A}{2})] = \lim_{x \to 0} [\frac{2 \sin^{2} (\frac{x}{2})}{2 \sin \frac{x}{2} \cos \frac{x}{2}}] = \lim_{x \to 0} (\tan (\frac{x}{2})) = 0$

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Question 60:

Evaluate the following limits:

$\lim_{x \to 0} \frac{x \{\sin (α + β) x + \sin (α - β) x + \sin 2 α x\}}{\cos^{2} β x - \cos^{2} α x}$

Answer:

Ans

Page No 29.51:

Question 61:

Evaluate the following limits:

$\lim_{x \to 0} \frac{\cos a x - \cos b x}{\cos c x - 1}$

Answer:

$\lim_{x \to 0} \frac{\cos a x - \cos b x}{\cos c x - 1} = \lim_{x \to 0} \frac{- 2 \sin (\frac{a x + b x}{2}) \sin (\frac{a x - b x}{2})}{- 2 \sin^{2} \frac{c x}{2}} (1 - \cos 2 θ = 2 \sin^{2} θ) = \lim_{x \to 0} \frac{(\frac{a + b}{2}) x \times \frac{\sin (\frac{a + b}{2}) x}{(\frac{a + b}{2}) x} \times (\frac{a - b}{2}) x \times \frac{\sin (\frac{a - b}{2}) x}{(\frac{a - b}{2}) x}}{\frac{c^{2} x^{2}}{4} \times \frac{\sin^{2} \frac{c x}{2}}{\frac{c^{2} x^{2}}{4}}} = \frac{(a + b) (a - b)}{c^{2}} \times \frac{\lim_{x \to 0} \frac{\sin (\frac{a + b}{2}) x}{(\frac{a + b}{2}) x} \times \lim_{x \to 0} \frac{\sin (\frac{a - b}{2}) x}{(\frac{a - b}{2}) x}}{{(\lim_{x \to 0} \frac{\sin \frac{c x}{2}}{\frac{c x}{2}})}^{2}}$
$= \frac{a^{2} - b^{2}}{c^{2}} \times \frac{1 \times 1}{1} (\lim_{θ \to 0} \frac{\sin θ}{θ} = 1) = \frac{a^{2} - b^{2}}{c^{2}}$

Page No 29.51:

Question 62:

Evaluate the following limits:

$\lim_{h \to 0} \frac{{(a + h)}^{2} \sin (a + h) - a^{2} \sin a}{h}$

Answer:

$\lim_{h \to 0} \frac{{(a + h)}^{2} \sin (a + h) - a^{2} \sin a}{h} = \lim_{h \to 0} \frac{(a^{2} + 2 a h + h^{2}) \sin (a + h) - a^{2} \sin a}{h} = \lim_{h \to 0} \frac{(2 a h + h^{2}) \sin (a + h) + a^{2} \sin (a + h) - a^{2} \sin a}{h} = \lim_{h \to 0} \frac{(2 a h + h^{2}) \sin (a + h)}{h} + \lim_{h \to 0} \frac{a^{2} \sin (a + h) - a^{2} \sin a}{h} = \lim_{h \to 0} (2 a + h) \sin (a + h) + a^{2} \lim_{h \to 0} \frac{\sin (a + h) - \sin a}{h}$
$= (2 a + 0) \sin (a + 0) + a^{2} \lim_{h \to 0} \frac{2 \sin (\frac{h}{2}) \cos (\frac{2 a + h}{2})}{h} = 2 a \sin a + a^{2} \times \lim_{h \to 0} \frac{\sin (\frac{h}{2})}{\frac{h}{2}} \times \lim_{h \to 0} \cos (\frac{2 a + h}{2}) = 2 a \sin a + a^{2} \times 1 \times \cos (\frac{2 a + 0}{2}) (\lim_{θ \to 0} \frac{\sin θ}{θ} = 1) = 2 a \sin a + a^{2} \cos a$

Page No 29.51:

Question 63:

If $\lim_{x \to 0} k x cosec x = \lim_{x \to 0} x cosec k x,$ find k.

Answer:

$\lim_{x \to 0} k x . cosec x = \lim_{x \to 0} x cosec k x \Rightarrow \lim_{x \to 0} [\frac{k x}{\sin x}] = \lim_{x \to 0} [\frac{x}{\sin (k x)}] \Rightarrow k \lim_{x \to 0} [\frac{x}{\sin x}] = \lim_{x \to 0} [\frac{k x}{\sin (k x)}] \times \frac{1}{k} \Rightarrow k = \frac{1}{k} \Rightarrow k^{2} = 1 \Rightarrow k = \pm 1$

Page No 29.62:

Question 1:

$\lim_{x \to π} \frac{\sin x}{π - x}$

Answer:

$\lim_{x \to π} \frac{\sin x}{π - x} = \lim_{h \to 0} \frac{\sin (π - h)}{π - (π - h)} [∵ \lim_{x \to a} f (x) = \lim_{h \to 0} f (a - h)] = \lim_{h \to 0} \frac{\sin h}{h} [∵ \sin (π - 0) = \sin 0] \Rightarrow 1$

Page No 29.62:

Question 2:

$\lim_{x \to \frac{π}{2}} \frac{\sin 2 x}{\cos x}$

Answer:

$\lim_{x \to \frac{π}{2}} \frac{\sin 2 x}{\cos x} [\sin 2 x = 2 \sin x \cos x] = \lim_{x \to \frac{π}{2}} \frac{2 \sin x \cos x}{\cos x} = \lim_{x \to \frac{π}{2}} 2 \sin x \Rightarrow 2$

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Question 3:

$\lim_{x \to \frac{π}{2}} \frac{\cos^{2} x}{1 - \sin x}$

Answer:

$\lim_{x \to \frac{π}{2}} \frac{\cos^{2} x}{1 - \sin x} = \lim_{x \to \frac{π}{2}} \frac{1 - \sin^{2} x}{1 - \sin x} [\begin{matrix} ∵ \cos^{2} x = 1 - \sin^{2} x \\ a^{2} - b^{2} = (a - b) (a + b) \end{matrix}] = \lim_{x \to \frac{π}{2}} \frac{(1 - \sin x) (1 + \sin x)}{(1 - \sin x)} = \lim_{x \to \frac{π}{2}} (1 + \sin x) \Rightarrow 1 + 1 = 2$

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Question 4:

Evaluate the following limits:

$\lim_{x \to \frac{π}{3}} \frac{\sqrt{1 - \cos 6 x}}{\sqrt{2} (\frac{π}{3} - x)}$

Answer:

$\lim_{x \to \frac{π}{3}} \frac{\sqrt{1 - \cos 6 x}}{\sqrt{2} (\frac{π}{3} - x)} = \lim_{x \to \frac{π}{3}} \frac{\sqrt{2 \sin^{2} 3 x}}{\sqrt{2} (\frac{π}{3} - x)} (1 - \cos 2 θ = 2 \sin^{2} θ) = \lim_{x \to \frac{π}{3}} \frac{\sqrt{2} \sin 3 x}{\sqrt{2} (\frac{π}{3} - x)} = \lim_{x \to \frac{π}{3}} \frac{\sin 3 x}{(\frac{π}{3} - x)}$
$= \lim_{h \to 0} \frac{\sin 3 (\frac{π}{3} + h)}{\frac{π}{3} - (\frac{π}{3} + h)} (Put x = \frac{π}{3} + h) = \lim_{h \to 0} \frac{\sin (π + 3 h)}{- h} = \lim_{h \to 0} \frac{- \sin 3 h}{- h} [\sin (π + θ) = - \sin θ] = 3 \times \lim_{h \to 0} \frac{\sin 3 h}{3 h} = 3 \times 1 (\lim_{θ \to 0} \frac{\sin θ}{θ} = 1) = 3$

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Question 5:

$\lim_{x \to a} \frac{\cos x - \cos a}{x - a}$

Answer:

$\lim_{x \to a} \frac{\cos x - \cos a}{x - a} = \lim_{x \to a} \frac{- 2 \sin (\frac{x + a}{2}) \sin (\frac{x - a}{2})}{2 (\frac{x - a}{2})} [∵ \cos A - \cos B - 2 \sin (\frac{A - B}{2}) \sin (\frac{A + B}{2})] = \lim_{x \to a} - \sin (\frac{x + a}{2}) [∵ \lim_{θ \to a} \sin \frac{(θ - a)}{(θ - a)} = 1] = - \sin (\frac{2 a}{2}) \Rightarrow - \sin a$

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Question 6:

$\lim_{x \to \frac{π}{4}} \frac{1 - \tan x}{x - \frac{π}{4}}$

Answer:

$\lim_{x \to \frac{π}{4}} \frac{1 - \tan x}{x - \frac{π}{4}} = \lim_{h \to 0} \frac{1 - \tan (\frac{π}{4} + h)}{(\frac{π}{4} + h - \frac{π}{4})} = \lim_{h \to 0} \frac{1 - (\frac{1 + \tan h}{1 - \tan h})}{h} [∵ \tan (\frac{π}{4} + 0) = \frac{1 + \tan θ}{1 - \tan θ}] = \lim_{h \to 0} \frac{1 - \tan h - 1 - \tan h}{(1 - \tan h) h} = \lim_{h \to 0} \frac{- 2 \tan h}{h (1 - \tan h)} [∵ \lim_{h \to 0} \frac{\tan h}{h} = 1] \Rightarrow \frac{- 2}{1 - 0} = - 2$

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Question 7:

$\lim_{x \to \frac{π}{2}} \frac{1 - \sin x}{{(\frac{π}{2} - x)}^{2}}$

Answer:

$\lim_{x \to \frac{π}{2}} \frac{1 - \sin x}{{(\frac{π}{2} - x)}^{2}} = \lim_{h \to 0} \frac{1 - \sin (\frac{π}{2} - h)}{{(\frac{π}{2} - (\frac{π}{2} - h))}^{2}} = \lim_{h \to 0} \frac{1 - \cos h}{h^{2}} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{\frac{4 h^{2}}{4}} [∵ \lim_{h \to 0} \frac{\sin h}{h} = 1] = \frac{1}{2} \lim_{h \to 0} {(\frac{\sin \frac{h}{2}}{\frac{h}{2}})}^{2} \Rightarrow \frac{1}{2}$

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Question 8:

$\lim_{x \to \frac{π}{3}} \frac{\sqrt{3} - \tan x}{π - 3 x}$

Answer:

$\lim_{x \to \frac{π}{3}} \frac{\sqrt{3} - \tan x}{π - 3 x} = \lim_{h \to 0} \frac{\sqrt{3} - \tan (\frac{π}{3} - h)}{π - 3 (\frac{π}{3} - h)} = \lim_{h \to 0} \frac{\sqrt{3} - (\frac{\tan \frac{π}{3} - \tan h}{1 + \tan \frac{π}{3} \tan h})}{π - 3 (\frac{π}{3} - h)} = \lim_{h \to 0} \frac{\sqrt{3} - (\frac{\sqrt{3} - \tan h}{1 + \sqrt{3} \tan h})}{3 h} = \lim_{h \to 0} \frac{\sqrt{3} + 3 \tan h - \sqrt{3} + \tan h}{(1 + \sqrt{3} \tan h) 3 h} = \lim_{h \to 0} \frac{4 \tan h}{3 h (1 + \sqrt{3} \tan h)} = \frac{4}{3}$

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Question 9:

$\lim_{x \to \frac{π}{2}} \frac{\cot x}{\frac{π}{2} - x}$

Answer:

$\lim_{x \to \frac{π}{2}} \frac{\cot x}{\frac{π}{2} - x} = \lim_{h \to 0} \frac{\cot (\frac{π}{2} - h)}{\frac{π}{2} - (\frac{π}{2} - h)} \Rightarrow \lim_{x \to 0} \frac{\tan h}{h} = 1$

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Question 10:

$\lim_{x \to \frac{π}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{π}{4}}$

Answer:

$\lim_{x \to \frac{π}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{π}{4}} Rationalising the numerator : = \lim_{x \to \frac{π}{4}} (\frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{π}{4}}) (\frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}}) = \lim_{x \to \frac{π}{4}} \frac{\cos x - \sin x}{(x - \frac{π}{4}) (\sqrt{\cos x} + \sqrt{\sin x})} = \lim_{h \to 0} \frac{\cos (\frac{π}{4} + h) - \sin (\frac{π}{4} + h)}{(\frac{π}{4} + h - \frac{π}{4}) (\sqrt{\cos (\frac{π}{4} + h)} + \sqrt{\sin (\frac{π}{4} + h)})} = \lim_{h \to 0} \frac{\cos \frac{π}{4} \cos h - \sin \frac{π}{4} \sin h - \sin \frac{π}{4} \cos h - \cos \frac{π}{4} \sin h}{h (\sqrt{\cos (\frac{π}{4} + h)} + \sqrt{\sin (\frac{π}{4} + h)})} = \lim_{h \to 0} \frac{- \sqrt{2} \sin h}{h (\sqrt{\cos (\frac{π}{4} + h)} + \sqrt{\sin (\frac{π}{4} + h)})} \Rightarrow \frac{- \sqrt{2} \times 1}{2 {(\frac{1}{2})}^{\frac{1}{4}}} = \frac{- 1}{2^{\frac{1}{4}}} = - 2^{- \frac{1}{4}}$

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Question 11:

$\lim_{x \to \frac{π}{2}} \frac{\sqrt{2 - \sin x} - 1}{{(\frac{π}{2} - x)}^{2}}$

Answer:

$\lim_{x \to \frac{π}{2}} \frac{\sqrt{2 - \sin x} - 1}{{(\frac{π}{2} - x)}^{2}} = \lim_{h \to 0} \frac{\sqrt{2 - \sin (\frac{π}{2} - h)} - 1}{{(\frac{π}{2} - (\frac{π}{2} - h))}^{2}} = \lim_{h \to 0} \frac{\sqrt{2 - \cos h} - 1}{h^{2}} Dividing the numerator and the denominator by \sqrt{2 - \cos h} + 1 : \lim_{h \to 0} \frac{(\sqrt{2 - \cos h} - 1) (\sqrt{2 - \cos h} + 1)}{(\sqrt{2 - \cos h} + 1) h^{2}} = \lim_{h \to 0} \frac{2 - \cos h - 1}{h^{2} (\sqrt{2 - \cos h} + 1)} = \lim_{h \to 0} \frac{1 - \cos h}{h^{2} [\sqrt{2 - \cos h} + 1]} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{\frac{4 h^{2}}{4} (\sqrt{2 - \cos h} + 1)} = \frac{1}{2} \lim_{h \to 0} {(\frac{\sin \frac{h}{2}}{\frac{h}{2}})}^{2} \times \lim_{h \to 0} (\frac{1}{\sqrt{2 - \cos h} + 1}) = \frac{1}{2 (\sqrt{2 - 1} + 1)} = \frac{1}{4}$

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Question 12:

$\lim_{x \to \frac{π}{4}} \frac{\sqrt{2} - \cos x - \sin x}{{(\frac{π}{4} - x)}^{2}}$

Answer:

$\lim_{x \to \frac{π}{4}} \frac{\sqrt{2} - \cos x - \sin x}{{(\frac{π}{4} - x)}^{2}} Dividing the numerator and the denominator by \sqrt{2} : \lim_{x \to \frac{π}{4}} \frac{1 - (\frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x)}{\frac{1}{\sqrt{2}} {(\frac{π}{4} - x)}^{2}} = \lim_{x \to \frac{π}{4}} \frac{1 - (\cos \frac{π}{4} \cos x + \sin \frac{π}{4} \sin x)}{\frac{1}{\sqrt{2}} {(\frac{π}{4} - x)}^{2}} = \lim_{x \to \frac{π}{4}} \frac{\sqrt{2} (1 - \cos (\frac{π}{4} - x))}{{(\frac{π}{4} - x)}^{2}} = \lim_{x \to \frac{π}{4}} \frac{\sqrt{2} (2 \sin^{2} (\frac{\frac{π}{4} - x}{2}))}{{(\frac{π}{4} - x)}^{2}} [∵ 1 - \cos θ = 2 \sin \frac{2 θ}{2}] = 2 \sqrt{2} \lim_{x \to \frac{π}{4}} \frac{[\sin^{2} (\frac{\frac{π}{4} - x}{2})]}{4 {(\frac{\frac{π}{4} - x}{2})}^{2}} = \frac{2 \sqrt{2}}{4} = \frac{1}{\sqrt{2}}$

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Question 13:

$\lim_{x \to \frac{π}{8}} \frac{\cot 4 x - \cos 4 x}{{(π - 8 x)}^{3}}$

Answer:

$\lim_{x \to \frac{π}{8}} \frac{\cot 4 x - \cos 4 x}{{(π - 8 x)}^{3}} = \lim_{h \to 0} [\frac{\cot 4 (\frac{π}{8} - h) - \cos 4 (\frac{π}{8} - h)}{{(π - 8 (\frac{π}{8} - h))}^{3}}] = \lim_{h \to 0} [\frac{\tan 4 h - \sin 4 h}{{(8 h)}^{3}}] = \lim_{h \to 0} [\frac{\sin 4 h - \cos 4 h \sin 4 h}{512 (\cos 4 h) h^{3}}] = \lim_{h \to 0} [\frac{\sin 4 h (1 - \cos 4 h)}{(\cos 4 h) 512 h^{3}}] = \lim_{h \to 0} [\frac{\tan 4 h}{4 h} \times \frac{2 \sin^{2} 2 h}{32 \times 4 h^{2}}] = \frac{1}{16} \lim_{h \to 0} (\frac{\tan 4 h}{4 h}) \times \lim_{h \to 0} {(\frac{\sin 2 h}{2 h})}^{2} = \frac{1}{16} \times 1 \times 1 = \frac{1}{16}$

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Question 14:

$\lim_{x \to a} \frac{\cos x - \cos a}{\sqrt{x} - \sqrt{a}}$

Answer:

$\lim_{x \to a} \frac{\cos x - \cos a}{\sqrt{x} - \sqrt{a}} = \lim_{x \to a} \frac{- 2 \sin (\frac{x + a}{2}) \sin (\frac{x - a}{2})}{\sqrt{x} - \sqrt{a}} Dividing the numerator and the denominator by \sqrt{x} + \sqrt{a} : - 2 \lim_{x \to a} \frac{(\sin (\frac{x + a}{2}) \sin (\frac{x - a}{2}))}{2 (\frac{x - a}{2})} (\sqrt{x} + \sqrt{a}) = - 2 \lim_{x \to a} \sin (\frac{x + a}{2}) \times \frac{\sin (\frac{x - a}{2})}{2 (\frac{x - a}{2})} (\sqrt{x} + \sqrt{a}) \Rightarrow - 2 \sin (\frac{2 a}{2}) (\frac{\sqrt{a} + \sqrt{a}}{2}) = - 2 \sqrt{a} \sin a$

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Question 15:

$\lim_{x \to π} \frac{\sqrt{5 + \cos x} - 2}{{(π - x)}^{2}}$

Answer:

$\lim_{x \to π} \frac{\sqrt{5 + \cos x} - 2}{{(π - x)}^{2}} = \lim_{h \to 0} \frac{\sqrt{5 + \cos (π - h)} - 2}{{(π - (π - h))}^{2}} = \lim_{h \to 0} \frac{\sqrt{5 - \cos h} - 2}{h^{2}} Rationalising the numerator, we get : \lim_{h \to 0} \frac{(\sqrt{5 - \cos h} - 2) (\sqrt{5 - \cos h} + 2)}{h^{2} (\sqrt{5 - \cos h} + 2)} = \lim_{h \to 0} \frac{5 - \cos h - 4}{h^{2} (\sqrt{5 - \cos h} + 2)} = \lim_{h \to 0} \frac{1 - \cos h}{h^{2} [\sqrt{5 - \cos h} + 2]} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{4 (\frac{h^{2}}{4}) (\sqrt{5 - \cos h} + 2)} = \frac{1}{2 (\sqrt{5 - 1} + 2)} = \frac{1}{2 (4)} = \frac{1}{8}$

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Question 16:

$\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a}$

Answer:

$\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a} = \lim_{x \to a} \frac{- 2 \sin (\frac{\sqrt{x} + \sqrt{a}}{2}) \sin (\frac{\sqrt{x} - \sqrt{a}}{2})}{2 (\frac{\sqrt{x} - \sqrt{a}}{2}) (\sqrt{x} + \sqrt{a})} = \lim_{x \to a} \frac{- \sin (\frac{\sqrt{x} + \sqrt{a}}{2})}{\sqrt{x} + \sqrt{a}} \times \frac{\sin (\frac{\sqrt{x} - \sqrt{a}}{2})}{(\frac{\sqrt{x} - \sqrt{a}}{2})} = \frac{- 1 \sin (\sqrt{a})}{2 \sqrt{a}} \times 1 = \frac{- \sin \sqrt{a}}{2 \sqrt{a}}$

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Question 17:

$\lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a}$

Answer:

$\lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a} = \lim_{x \to a} \frac{2 \cos (\frac{\sqrt{x} + \sqrt{a}}{2}) \sin (\frac{\sqrt{x} - \sqrt{a}}{2})}{(\sqrt{x} + \sqrt{a}) (\sqrt{x} - \sqrt{a})} = \lim_{x \to a} \frac{2 \cos (\frac{\sqrt{x} + \sqrt{a}}{2})}{\sqrt{x} + \sqrt{a}} \times \frac{\sin (\frac{\sqrt{x} - \sqrt{a}}{2})}{2 (\frac{\sqrt{x} - \sqrt{a}}{2})} = \frac{1}{\sqrt{a} + \sqrt{a}} \cos (\frac{2 \sqrt{a}}{2}) \Rightarrow \frac{1}{2 \sqrt{a}} \cos \sqrt{a}$

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Question 18:

$\lim_{x \to 1} \frac{1 - x^{2}}{\sin 2 π x}$

Answer:

$\lim_{x \to 1} \frac{1 - x^{2}}{\sin 2 π x} = \lim_{h \to 0} \frac{1 - {(1 - h)}^{2}}{\sin 2 π (1 - h)} = \lim_{h \to 0} \frac{2 h - h^{2}}{- \sin 2 π h} = \lim_{h \to 0} \frac{- (2 - h)}{\frac{\sin 2 π h}{h}} = \lim_{h \to 0} \frac{(h - 2)}{\frac{2 π \sin 2 π h}{(2 π h)}} = \frac{0 - 2}{2 π \times 1} = \frac{- 1}{π}$

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Question 19:

$\lim_{x \to \frac{π}{4}} \frac{f (x) - f (\frac{π}{4})}{x - \frac{π}{4}},$ where f(x) = sin 2x

Answer:

$\lim_{x \to \frac{π}{4}} \frac{f (x) - f (\frac{π}{4})}{x - \frac{π}{4}} = \lim_{h \to 0} \frac{f (\frac{π}{4} + h) - f (\frac{π}{4})}{\frac{π}{4} + h - \frac{π}{4}} It is given that f (x) = \sin 2 x . \Rightarrow \lim_{h \to 0} \frac{\sin (\frac{π}{2} + 2 h) - \sin (\frac{π}{2})}{h} = \lim_{h \to 0} \frac{\cos 2 h - 1}{h} = \lim_{h \to 0} - 2 (\frac{\sin^{2} h}{h \times h}) (h) \Rightarrow \lim_{h \to 0} (- 2 h) = 0$

Page No 29.62:

Question 20:

$\lim_{x \to 1} \frac{1 + \cos π x}{{(1 - x)}^{2}}$

Answer:

$\lim_{x \to 1} \frac{1 + \cos π x}{{(1 - x)}^{2}} = \lim_{h \to 0} \frac{1 + \cos π (1 - h)}{{(1 - (1 - h))}^{2}} = \lim_{h \to 0} \frac{1 - \cos π h}{h^{2}} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{π h}{2}}{\frac{4}{π^{2}} (\frac{π^{2} h^{2}}{4})} = \lim_{h \to 0} \frac{\sin^{2} \frac{π h}{2}}{\frac{2}{π^{2}} {(\frac{π h}{2})}^{2}} \Rightarrow \frac{π^{2}}{2}$

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Question 21:

$\lim_{x \to 1} \frac{1 - x^{2}}{\sin π x}$

Answer:

$\lim_{x \to 1} \frac{1 - x^{2}}{\sin π x} \lim_{h \to 0} \frac{1 - {(1 - h)}^{2}}{\sin π (1 - h)} \lim_{h \to 0} \frac{2 h - h^{2}}{\sin π h} = \lim_{h \to 0} \frac{h (2 - h)}{\sin π h} = \lim_{h \to 0} \frac{(2 - h)}{π \times \frac{\sin π h}{π h}} \Rightarrow \frac{2 - 0}{π} = \frac{2}{π}$

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Question 22:

$\lim_{x \to \frac{π}{4}} \frac{1 - \sin 2 x}{1 + \cos 4 x}$

Answer:

$\lim_{x \to \frac{π}{4}} \frac{1 - \sin 2 x}{1 + \cos 4 x} = \lim_{h \to 0} \frac{1 - \sin 2 (\frac{π}{4} - h)}{1 + \cos 4 (\frac{π}{4} - h)} = \lim_{h \to 0} \frac{1 - \cos 2 h}{1 - \cos 4 h} = \lim_{h \to 0} \frac{2 \sin^{2} h}{2 \sin^{2} 2 h} \Rightarrow \lim_{h \to 0} \frac{\frac{\sin^{2} h}{h^{2}}}{\frac{4 \sin^{2} 2 h}{4 h^{2}}} \Rightarrow \frac{1}{4}$

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Question 23:

$\lim_{x \to 1} \frac{1 - \frac{1}{x}}{\sin π (x - 1)}$

Answer:

$\lim_{x \to 1} [\frac{1 - \frac{1}{x}}{sinπ (x - 1)}] = \lim_{x \to 1} [\frac{x - 1}{x sinπ (x - 1)}]$

Let y = x – 1
If x → 1, then y → 0.

$= \lim_{y \to 0} [\frac{y}{(y + 1) sinπ y}] = \lim_{y \to 0} [\frac{1}{π (y + 1) \frac{\sin πy}{πy}}] = \frac{1}{π (0 + 1) \times 1} = \frac{1}{π}$

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Question 24:

$\lim_{n \to \infty} n \sin (\frac{π}{4 n}) \cos (\frac{π}{4 n})$

Answer:

$\lim_{n \to \infty} n \sin \frac{π}{4 n} \cos \frac{π}{4 n} = \lim_{n \to \infty} n \sin \frac{π}{4 n} \times \lim_{n \to \infty} \cos \frac{π}{4 n} = \lim_{n \to \infty} \frac{n \sin \frac{π}{4 n}}{\frac{π}{4 n}} \times \frac{π}{4 n} \times 1 = \frac{π}{4} \lim_{n \to \infty} \frac{\sin \frac{π}{4 n}}{\frac{π}{4 n}} Let y = \frac{π}{4 n} If n \to \infty, then y \to 0 . = \frac{π}{4} \lim_{y \to 0} \frac{\sin y}{y} = \frac{π}{4}$

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Question 25:

$\lim_{n \to \infty} 2^{n - 1} \sin (\frac{a}{2^{n}})$

Answer:

$\lim_{n \to \infty} 2^{n - 1} \sin (\frac{a}{2^{n}}) = \lim_{n \to \infty} \frac{2^{n}}{2} \times \frac{\sin (\frac{a}{2^{n}})}{(\frac{a}{2^{n}})} \times (\frac{a}{2^{n}}) Let y = \frac{a}{2^{n}} If n \to \infty, then y \to 0 . = \lim_{y \to 0} \frac{a}{2} \times (\frac{\sin y}{y}) \Rightarrow \frac{a}{2}$

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Question 26:

$\lim_{n \to \infty} \frac{\sin (\frac{a}{2^{n}})}{\sin (\frac{b}{2^{n}})}$

Answer:

$\lim_{n \to \infty} \frac{\sin (\frac{a}{2^{n}})}{\sin (\frac{b}{2^{n}})} = \lim_{n \to \infty} \frac{(\frac{a}{2^{n}}) \sin (\frac{a}{2^{n}})}{(\frac{a}{2^{n}}) \times (\frac{b}{2^{n}}) \times (\frac{\sin (\frac{b}{2^{n}})}{\frac{b}{2^{n}}})} Let : y = \frac{a}{2^{n}} z = \frac{b}{2^{n}} If n \to \infty, then y \to 0 and z \to 0 . = \frac{y}{z} \lim_{y \to 0} \frac{\sin y}{y} \times \frac{1}{\lim_{z \to 0} \frac{\sin z}{z}} = \frac{y}{z} \times 1 \times \frac{1}{1} = \frac{\frac{a}{2^{n}} \times 1}{\frac{b}{2^{n}} \times 1} = \frac{a}{b}$

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Question 27:

$\lim_{x \to - 1} \frac{x^{2} - x - 2}{(x^{2} + x) + \sin (x + 1)}$

Answer:

$\lim_{x \to - 1} \frac{x^{2} - x - 2}{x^{2} + x + \sin (x + 1)} = \lim_{x \to - 1} \frac{(x - 2) (x + 1)}{x (x + 1) + \sin (x + 1)} Let y = x + 1 If x \to - 1, then y \to 0 . = \lim_{y \to 0} \frac{(y - 3) y}{(y - 1) y + \sin y} Dividing the numerator and the denominator by y : = \lim_{y \to 0} \frac{(y - 3)}{(y - 1) + \frac{\sin y}{y}} = \frac{0 - 3}{(0 - 1) + 1} \Rightarrow \frac{- 3}{0} = \infty$

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Question 28:

$\lim_{x \to 2} \frac{x^{2} - x - 2}{x^{2} - 2 x + \sin (x - 2)}$

Answer:

$\lim_{x \to 2} \frac{x^{2} - x - 2}{x^{2} - 2 x + \sin (x - 2)} = \lim_{x \to 2} \frac{(x - 2) (x + 1)}{x (x - 2) + \sin (x - 2)} Let y = x - 2 x \to 2 ∴ y \to 0 = \lim_{y \to 0} \frac{y (y + 3)}{(y + 2) y + \sin y} Dividing the numerator and the denominator by y: = \lim_{y \to 0} \frac{(y + 3)}{(y + 2) + \frac{\sin y}{y}} = \frac{3}{2 + 1} = \frac{3}{3} = 1$

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Question 29:

$\lim_{x \to 1} (1 - x) \tan (\frac{π x}{2})$

Answer:

$\lim_{x \to 1} (1 - x) \tan \frac{π x}{2} = \lim_{h \to 0} \{1 - (1 - h)\} \tan \frac{π}{2} (1 - h) = \lim_{h \to 0} h \tan (\frac{π}{2} - \frac{π h}{2}) = \lim_{h \to 0} h \cot \frac{π h}{2} = \lim_{h \to 0} \frac{h}{\tan \frac{π h}{2}} = \lim_{h \to 0} \frac{1}{\frac{\tan \frac{π h}{2} \times \frac{π}{2}}{\frac{πh}{2}}} = \frac{1}{\frac{π}{2}} [∵ \lim_{h \to 0} \tan \frac{h}{h} = 1] = \frac{2}{π}$

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Question 30:

$\lim_{x \to \frac{π}{4}} \frac{1 - \tan x}{1 - \sqrt{2} \sin x}$

Answer:

$\lim_{x \to \frac{π}{4}} [\frac{1 - \tan x}{1 - \sqrt{2} \sin x}] It is of \frac{0}{0} form .$

Rationalising the denominator, we get:

$\lim_{x \to \frac{π}{4}} [\frac{(1 - \tan x) (1 + \sqrt{2} \sin x)}{(1 - \sqrt{2} \sin x) (1 + \sqrt{2} \sin x)}] = \lim_{x \to \frac{π}{4}} [\frac{(1 - \tan x) (1 + \sqrt{2} \sin x)}{1 - 2 \sin^{2} x}] = \lim_{x \to \frac{π}{4}} [\frac{(1 - \frac{\sin x}{\cos x}) (1 + \sqrt{2} \sin x)}{\cos 2 x}] = \lim_{x \to \frac{π}{4}} [\frac{(\cos x - \sin x) (1 + \sqrt{2} \sin x)}{\cos x \cos 2 x}] = \lim_{x \to \frac{π}{4}} [\frac{(\cos x - \sin x) (1 + \sqrt{2} \sin x)}{\cos x \cdot (\cos^{2} x - \sin^{2} x)}] = \lim_{x \to \frac{π}{4}} [\frac{(\cos x - \sin x) (1 + \sqrt{2} \sin x)}{\cos x [\cos x - \sin x] [\cos x + \sin x]}] = \frac{(1 + \sqrt{2} \times \frac{1}{\sqrt{2}})}{(\frac{1}{\sqrt{2}}) (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})} = \frac{2}{\frac{1}{\sqrt{2}} \times \sqrt{2}} = 2$

Page No 29.63:

Question 31:

$\lim_{x \to π} \frac{\sqrt{2 + \cos x} - 1}{{(π - x)}^{2}}$

Answer:

$\lim_{x \to π} \frac{\sqrt{2 + \cos x} - 1}{{(π - x)}^{2}} = \lim_{h \to 0} \frac{\sqrt{2 + \cos (π - h)} - 1}{{(π - (π - h))}^{2}} = \lim_{h \to 0} (\frac{\sqrt{2 - \cos h} - 1}{h^{2}}) \times \frac{\sqrt{2 - \cos h} + 1}{\sqrt{2 - \cos h} + 1} = \lim_{h \to 0} \frac{2 - \cos h - 1}{h^{2} (\sqrt{2 - \cos h} + 1)} = \lim_{h \to 0} \frac{1 - \cos h}{h^{2} (\sqrt{2 - \cos h} + 1)} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{4 \times \frac{h^{2}}{4} \times [\sqrt{2 - \cos h} + 1]} = \frac{1 \times 1^{2}}{2 (\sqrt{2 - 1} + 1)} = \frac{1}{4}$

Page No 29.63:

Question 32:

$\lim_{x \to π} \frac{1 + \cos x}{\tan^{2} x}$

Answer:

$\lim_{x \to π} \frac{1 + \cos x}{\tan^{2} x} = \lim_{h \to 0} \frac{1 + \cos (π + h)}{\tan^{2} (π + h)} = \lim_{h \to 0} \frac{1 - \cos h}{\tan^{2} h} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{\tan^{2} h} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{4 (\frac{h^{2}}{4}) \times \frac{\tan^{2} h}{h^{2}}} = \frac{1}{2} \frac{\lim_{h \to 0} {(\frac{\sin \frac{h}{2}}{\frac{h}{2}})}^{2}}{{\lim_{h \to 0} (\frac{\tan h}{h})}^{2}} = \frac{1}{2} \times \frac{{(1)}^{2}}{{(1)}^{2}} = \frac{1}{2}$

Page No 29.63:

Question 33:

$\lim_{x \to \frac{π}{2}} (\frac{π}{2} - x) \tan x$

Answer:

$\lim_{x \to 1} \frac{1 - \frac{1}{x}}{\sin π (x - 1)} = \lim_{x \to 1} \frac{x - 1}{x \sin π (x - 1)} Let y = x - 1 x \to 1 ∴ y \to 0 = \lim_{y \to 0} \frac{y}{(y + 1) \sin π y} = \lim_{y \to 0} \frac{1}{π (y + 1) \times \frac{\sin π y}{π y}} = \frac{1}{π (0 + 1) \times 1} = \frac{1}{π}$

Page No 29.63:

Question 34:

$\lim_{x \to \frac{π}{6}} \frac{\cot^{2} x - 3}{cosec x - 2}$

Answer:

$\lim_{x \to \frac{π}{6}} [\frac{\cot^{2} x - 3}{cosec x - 2}] = \lim_{x \to \frac{π}{6}} [\frac{{cosec}^{2} x - 1 - 3}{cosec x - 2}] = \lim_{x \to \frac{π}{6}} [\frac{{cosec}^{2} x - 4}{cosec x - 2}] = \lim_{x \to \frac{π}{6}} [\frac{(cosec x - 2) (cosec x + 2)}{(cosec x - 2)}] = cosec \frac{π}{6} + 2 = 4$

Page No 29.63:

Question 35:

$\lim_{x \to \frac{π}{4}} \frac{\sqrt{2} - \cos x - \sin x}{{(4 x - π)}^{2}}$

Answer:

$\lim_{x \to \frac{π}{4}} \frac{\sqrt{2} - \cos x - \sin x}{{(4 x - π)}^{2}} = \lim_{h \to 0} \frac{\sqrt{2} - \{\cos (\frac{π}{4} + h) + \sin (\frac{π}{4} + h)\}}{{(4 (\frac{π}{4} + h) - π)}^{2}} = \lim_{h \to 0} \frac{\sqrt{2} - \{\cos \frac{π}{4} \cos h - \sin \frac{π}{4} \sin h + \sin \frac{π}{4} \cos h + \cos \frac{π}{4} \sin h\}}{{(4 h)}^{2}} = \lim_{h \to 0} \frac{\sqrt{2} - \sqrt{2} \cos h}{{(4 h)}^{2}} = \lim_{h \to 0} \frac{\sqrt{2} (1 - \cos h)}{16 h^{2}} = \lim_{h \to 0} \frac{2 \sqrt{2} \sin^{2} \frac{h}{2}}{64 \times \frac{h^{2}}{4}} = \frac{\sqrt{2}}{32} \times {(1)}^{2} = \frac{1}{16 \sqrt{2}}$

Page No 29.63:

Question 36:

$\lim_{x \to \frac{π}{2}} \frac{(\frac{π}{2} - x) \sin x - 2 \cos x}{(\frac{π}{2} - x) + \cot x}$

Answer:

$\lim_{x \to \frac{π}{2}} \frac{(\frac{π}{2} - x) \sin x - 2 \cos x}{(\frac{π}{2} - x) + \cot x} = \lim_{h \to 0} \frac{[\frac{π}{2} - (\frac{π}{2} - h)] \sin (\frac{π}{2} - h) - 2 \cos (\frac{π}{2} - h)}{[\frac{π}{2} - (\frac{π}{2} - h)] + \cot (\frac{π}{2} - h)} (Plugging x = \frac{π}{2} - h) = \lim_{h \to 0} \frac{h \cos h - 2 \sin h}{h + \tan h} Dividing the numerator and the denominator by h: \lim_{h \to 0} \frac{\cos h - 2 \frac{\sin h}{h}}{1 + \frac{\tan h}{h}} \Rightarrow \frac{1 - 2 (1)}{1 + 1} = \frac{- 1}{2}$

Page No 29.63:

Question 37:

$\lim_{x \to \frac{π}{4}} \frac{\cos x - \sin x}{(\frac{π}{4} - x) (\cos x + \sin x)}$

Answer:

$\lim_{x \to \frac{π}{4}} \frac{\cos x - \sin x}{(\frac{π}{4} - x) (\cos x + \sin x)} Dividing the numerator and the denominator by \sqrt{2} : \lim_{x \to \frac{π}{4}} \frac{\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x}{(\frac{π}{4} - x) \frac{(\cos x + \sin x)}{\sqrt{2}}} = \lim_{x \to \frac{π}{4}} \frac{\sqrt{2} (\sin \frac{π}{4} \cos x - \cos \frac{π}{4} \sin x)}{(\frac{π}{4} - x) (\cos x + \sin x)} = \lim_{x \to \frac{π}{4}} \frac{\sqrt{2} (\sin (\frac{π}{4} - x))}{(\frac{π}{4} - x) (\cos x + \sin x)} = \lim_{x \to \frac{π}{4}} \frac{\sqrt{2}}{\sin x + \cos x} \times \lim_{x \to \frac{π}{4}} \frac{\sin (\frac{π}{4} - x)}{(\frac{π}{4} - x)} \Rightarrow \frac{\sqrt{2}}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} \times 1 = 1$

Page No 29.63:

Question 38:

Evaluate the following limits:

$\lim_{x \to π} \frac{1 - \sin \frac{x}{2}}{\cos \frac{x}{2} (\cos \frac{x}{4} - \sin \frac{x}{4})}$ [NCERT EXEMPLAR]

Answer:

$\lim_{x \to π} \frac{1 - \sin \frac{x}{2}}{\cos \frac{x}{2} (\cos \frac{x}{4} - \sin \frac{x}{4})}$

Put $x = π + h$

When $x \to π, h \to 0$

$∴ \lim_{x \to π} \frac{1 - \sin \frac{x}{2}}{\cos \frac{x}{2} (\cos \frac{x}{4} - \sin \frac{x}{4})} = \lim_{h \to 0} \frac{1 - \sin (\frac{π + h}{2})}{\cos (\frac{π + h}{2}) [\cos (\frac{π + h}{4}) - \sin (\frac{π + h}{4})]} = \lim_{h \to 0} \frac{1 - \sin (\frac{π}{2} + \frac{h}{2})}{\cos (\frac{π}{2} + \frac{h}{2}) [\cos (\frac{π}{4} + \frac{h}{4}) - \sin (\frac{π}{4} + \frac{h}{4})]} = \lim_{h \to 0} \frac{1 - \cos (\frac{h}{2})}{- \sin (\frac{h}{2}) [(\cos \frac{π}{4} \cos \frac{h}{4} - \sin \frac{π}{4} \sin \frac{h}{4}) - (\sin \frac{π}{4} \cos \frac{h}{4} + \cos \frac{π}{4} \sin \frac{h}{4})]}$
$= \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{4}}{- 2 \sin \frac{h}{4} \cos \frac{h}{4} (\frac{1}{\sqrt{2}} \cos \frac{h}{4} - \frac{1}{\sqrt{2}} \sin \frac{h}{4} - \frac{1}{\sqrt{2}} \cos \frac{h}{4} - \frac{1}{\sqrt{2}} \sin \frac{h}{4})} = \lim_{h \to 0} \frac{\sin \frac{h}{4}}{- \cos \frac{h}{4} \times (- \sqrt{2} \sin \frac{h}{4})} = \frac{1}{\sqrt{2}} \times \frac{1}{\lim_{h \to 0} \cos \frac{h}{4}} = \frac{1}{\sqrt{2}} \times 1 (\lim_{θ \to 0} \cos θ = 1) = \frac{1}{\sqrt{2}}$

Page No 29.65:

Question 1:

$\lim_{x \to π} \frac{1 + \cos x}{\tan^{2} x}$

Answer:

$\lim_{x \to π} (\frac{1 + \cos x}{\tan^{2} x}) [\frac{0}{0} form] = \lim_{x \to π} [\frac{(1 + \cos x)}{\sin^{2} x} \times \cos^{2} x] = \lim_{x \to π} [\frac{(1 + \cos x)}{1 - \cos^{2} x} \times \cos^{2} x] = \lim_{x \to π} [\frac{(1 + \cos x)}{(1 - \cos x) (1 + \cos x)} \times \cos^{2} x] = \lim_{x \to π} [\frac{\cos^{2} x}{(1 - \cos x)}] = \frac{\cos^{2} π}{1 - \cos π} = \frac{{(- 1)}^{2}}{1 - (- 1)} = \frac{1}{2}$

Page No 29.65:

Question 2:

$\lim_{x \to \frac{π}{4}} \frac{{cosec}^{2} x - 2}{\cot x - 1}$

Answer:

$\lim_{x \to \frac{π}{4}} [\frac{{cosec}^{2} x - 2}{\cot x - 1}] = \lim_{x \to \frac{π}{4}} [\frac{1 + \cot^{2} x - 2}{\cot x - 1}] = \lim_{x \to \frac{π}{4}} [\frac{\cot^{2} x - 1}{\cot x - 1}] = \lim_{x \to \frac{π}{4}} [\frac{(\cot x - 1) (\cot x + 1)}{(\cot x - 1)}] = \cot \frac{π}{4} + 1 = 2$

Page No 29.65:

Question 3:

$\lim_{x \to \frac{π}{6}} \frac{\cot^{2} x - 3}{cosec x - 2}$

Answer:

$\lim_{x \to \frac{π}{6}} [\frac{\cot^{2} x - 3}{cosec x - 2}] = \lim_{x \to \frac{π}{6}} [\frac{{cosec}^{2} x - 1 - 3}{cosec x - 2}] = \lim_{x \to \frac{π}{6}} [\frac{{cosec}^{2} x - 4}{cosec x - 2}] = \lim_{x \to \frac{π}{6}} [\frac{(cosec x - 2) (cosec x + 2)}{(cosec x - 2)}] = cosec \frac{π}{6} + 2 = 2 + 2 = 4$

Page No 29.65:

Question 4:

$\lim_{x \to \frac{π}{4}} \frac{2 - {cosec}^{2} x}{1 - \cot x}$

Answer:

$\lim_{x \to \frac{π}{4}} [\frac{2 - {cosec}^{2} x}{1 - \cot x}] = \lim_{x \to \frac{π}{4}} [\frac{2 - (1 + \cot^{2} x)}{1 - \cot x}] = \lim_{x \to \frac{π}{4}} [\frac{1 - \cot^{2} x}{1 - \cot x}] = \lim_{x \to \frac{π}{4}} [\frac{(1 - \cot x) (1 + \cot x)}{(1 - \cot x)}] = 1 + \cot (\frac{π}{4}) = 1 + 1 = 2$

Page No 29.65:

Question 5:

$\lim_{x \to π} \frac{\sqrt{2 + \cos x} - 1}{{(π - x)}^{2}}$

Answer:

$\lim_{x \to π} [\frac{\sqrt{2 + \cos x} - 1}{{(π - x)}^{2}}] Rationalising the numerator, we get : \lim_{x \to π} [\frac{(\sqrt{2 + \cos x} - 1) \times (\sqrt{2 + \cos x} + 1)}{{(π - x)}^{2} (\sqrt{2 + \cos x} + 1)}] = \lim_{x \to π} [\frac{2 + \cos x - 1}{{(π - x)}^{2} (\sqrt{2 + \cos x} + 1)}] = \lim_{x \to π} [\frac{1 + \cos x}{{(π - x)}^{2} [\sqrt{2 + \cos x} + 1]}]$

Let x = π $-$ h
when x → π, then h → 0

$= \lim_{h \to 0} [\frac{1 + \cos (π - h)}{{[π - (π - h)]}^{2} [\sqrt{2 + \cos (π - h)} + 1]}] = \lim_{h \to 0} [\frac{1 - \cos h}{h^{2} [\sqrt{2 - \cos h} + 1]}] \{∵ \cos (π - θ) = - \cos θ\} = \lim_{h \to 0} [\frac{2 \sin^{2} (\frac{h}{2})}{4 \times \frac{h^{2}}{4} [\sqrt{2 - \cos h} + 1]}] = \frac{1}{2} \lim_{h \to 0} [{(\frac{\sin \frac{h}{2}}{\frac{h}{2}})}^{2} \times \frac{1}{[\sqrt{2 - \cos h} + 1]}] = \frac{1}{2} \times 1 \times \frac{1}{(\sqrt{2 - \cos 0} + 1)} = \frac{1}{2} \times \frac{1}{(\sqrt{1} + 1)} = \frac{1}{2 \times 2} = \frac{1}{4}$

Page No 29.65:

Question 6:

$\lim_{x \to \frac{3 π}{2}} \frac{1 + {cosec}^{3} x}{\cot^{2} x}$

Answer:

$\lim_{x \to \frac{3 π}{3}} [\frac{1 + {cosec}^{3} x}{\cot^{2} x}] = \lim_{x \to \frac{3 π}{2}} [\frac{(1 + cosec x) (1^{2} + {cosec}^{2} x - cosec x)}{({cosec}^{2} x - 1)}] = \lim_{x \to \frac{3 π}{2}} [\frac{(1 + cosec x) (1 + {cosec}^{2} x - cosec x)}{(cosec x - 1) (cosec x + 1)}] = \frac{1 + {cosec}^{2} (\frac{3 π}{2}) - cosec (\frac{3 π}{2})}{cosec (\frac{3 π}{2}) - 1} = \frac{1 + 1 + 1}{- 1 - 1} = - \frac{3}{2}$

Page No 29.71:

Question 1:

$\lim_{x \to 0} \frac{5^{x} - 1}{\sqrt{4 + x} - 2}$

Answer:

$\lim_{x \to 0} [\frac{5^{x} - 1}{\sqrt{4 + x} - 2}]$

Rationalising the denominator, we get:

$= \lim_{x \to 0} [\frac{(5^{x} - 1) (\sqrt{4 + x} + 2)}{(\sqrt{4 + x} - 2) (\sqrt{4 + x} + 2)}] = \lim_{x \to 0} [\frac{(5^{x} - 1) (\sqrt{4 + x} + 2)}{4 + x - 4}] = \lim_{x \to 0} [(\frac{5^{x} - 1}{x}) (\sqrt{4 + x} + 2)] \{∵ \lim_{x \to 0} (\frac{a^{x} - 1}{x}) = \log a\} = \log 5 \times (\sqrt{4 + 0} + 2) = 4 \log 5$

Page No 29.71:

Question 2:

$\lim_{x \to 0} \frac{\log (1 + x)}{3^{x} - 1}$

Answer:

$\lim_{x \to 0} [\frac{\log (1 + x)}{3^{x} - 1}]$

Dividing the numerator and the denominator by x:

$\lim_{x \to 0} [\frac{\log (1 + x)}{x \cdot (\frac{3^{x} - 1}{x})}] [∵ \lim_{x \to 0} \frac{\log (1 + x)}{x} = 1 \lim_{x \to 0} (\frac{a^{x} - 1}{x}) = \log a] = \frac{1}{\log 3}$

Page No 29.71:

Question 3:

$\lim_{x \to 0} \frac{a^{x} + a^{- x} - 2}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{a^{x} + a^{- x} - 2}{x^{2}}] = \lim_{x \to 0} [\frac{{(a^{\frac{x}{2}})}^{2} + {(a^{- \frac{x}{2}})}^{2} - 2 a^{\frac{x}{2}} \cdot a^{- \frac{x}{2}}}{x^{2}}] = \lim_{x \to 0} [\frac{{(a^{\frac{x}{2}} - a^{- \frac{x}{2}})}^{2}}{x^{2}}] = \lim_{x \to 0} [\frac{{(a^{\frac{x}{2}} - \frac{1}{a^{\frac{x}{2}}})}^{2}}{x^{2}}] = \lim_{x \to 0} [\frac{{(a^{x} - 1)}^{2}}{x^{2}} \times \frac{1}{{(a^{\frac{x}{2}})}^{2}}] = \lim_{x \to 0} [{(\frac{a^{x} - 1}{x})}^{2} \times \frac{1}{a^{x}}] = \frac{{(\log a)}^{2}}{a^{0}} = {(\log a)}^{2}$

Page No 29.71:

Question 4:

$\lim_{x \to 0} \frac{a^{m x} - 1}{b^{n x} - 1}, n \neq 0$

Answer:

$\lim_{x \to 0} (\frac{a^{m x} - 1}{b^{n x} - 1}) = \lim_{x \to 0} ((\frac{a^{m x} - 1}{m x}) \times \frac{m x}{(\frac{b^{n x} - 1}{n x}) \times n x}) = \frac{\log_{e} (a)}{\log_{e} (b)} \times \frac{m}{n} = \frac{m}{n} \frac{\log a}{\log b}$

Page No 29.71:

Question 5:

$\lim_{x \to 0} \frac{a^{x} + b^{x} - 2}{x}$

Answer:

$\lim_{x \to 0} [\frac{a^{x} + b^{x} - 2}{x}] = \lim_{x \to 0} [\frac{a^{x} - 1 + b^{x} - 1}{x}] = \lim_{x \to 0} [\frac{a^{x} - 1}{x}] + \lim_{x \to 0} [\frac{b^{x} - 1}{x}] = \log a + \log b = \log (a b)$

Page No 29.71:

Question 6:

$\lim_{x \to 0} \frac{9^{x} - 2 . 6^{x} + 4^{x}}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{9^{x} - 2 . 6^{x} + 4^{x}}{x^{2}}] = \lim_{x \to 0} [\frac{{(3^{x})}^{2} - 2 \cdot 3^{x} \cdot 2^{x} + {(2^{x})}^{2}}{x^{2}}] = \lim_{x \to 0} [\frac{{(3^{x} - 2^{x})}^{2}}{x^{2}}] = \lim_{x \to 0} [{(\frac{3^{x} - 2^{x}}{2^{x}})}^{2} \times \frac{{(2^{x})}^{2}}{x^{2}}] = \lim_{x \to 0} {[\frac{{(\frac{3}{2})}^{x} - 1}{x}]}^{2} \times 2^{2 x} = {[\log (\frac{3}{2})]}^{2} \times 2^{0} = {[\log (\frac{3}{2})]}^{2}$

Page No 29.71:

Question 7:

$\lim_{x \to 0} \frac{8^{x} - 4^{x} - 2^{x} + 1}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{8^{x} - 4^{x} - 2^{x} + 1}{x^{2}}] = \lim_{x \to 0} [\frac{{(2^{x})}^{3} - {(2^{x})}^{2} - 2^{x} + 1}{x^{2}}] = \lim_{x \to 0} [\frac{{(2^{x})}^{2} (2^{x} - 1) - 1 (2^{x} - 1)}{x^{2}}] = \lim_{x \to 0} [\frac{(2^{2 x} - 1) (2^{x} - 1)}{x^{2}}] = \lim_{x \to 0} [\frac{2 (2^{2 x} - 1)}{2 x} \times (\frac{2^{x} - 1}{x})] = 2 \log 2 \times \log 2 = \log {(2)}^{2} \times \log 2 = (\log 4) \times (\log 2)$

Page No 29.71:

Question 8:

$\lim_{x \to 0} \frac{a^{m x} - b^{n x}}{x}$

Answer:

$\lim_{x \to 0} [\frac{a^{m x} - b^{n x}}{x}] = \lim_{x \to 0} [\frac{a^{m x} - 1 - b^{n x} + 1}{x}] = \lim_{x \to 0} [\frac{(a^{m x} - 1) - (b^{n x} - 1)}{x}] = \lim_{x \to 0} [(\frac{a^{m x} - 1}{m x}) \times m - (\frac{b^{n x} - 1}{n x}) \times n] = m \log a - n \log b = \log {(a)}^{m} - \log {(b)}^{n} = \log (\frac{a^{m}}{b^{n}})$

Page No 29.71:

Question 9:

$\lim_{x \to 0} \frac{a^{x} + b^{x} + c^{x} - 3}{x}$

Answer:

$\lim_{x \to 0} [\frac{a^{n} + b^{n} + c^{n} - 3}{x}] = \lim_{x \to 0} [\frac{(a^{n} - 1)}{x} + \frac{(b^{n} - 1)}{x} + \frac{(c^{n} - 1)}{x}] = \log a + \log b + \log c = \log (a b c)$

Page No 29.71:

Question 10:

$\lim_{x \to 2} \frac{x - 2}{\log_{a} (x - 1)}$

Answer:

$\lim_{x \to 2} [\frac{x - 2}{\log_{a} (x - 1)}]$

Let x = 2 + h

x → 2
∴ h → 0

$= \lim_{h \to 0} [\frac{(2 + h) - 2}{\frac{\log \{(2 + h) - 1\}}{\log a}}] = \log a \lim_{h \to 0} [\frac{h}{\log (1 + h)}] = \log a \times 1$

Page No 29.71:

Question 11:

$\lim_{x \to 0} \frac{5^{x} + 3^{x} + 2^{x} - 3}{x}$

Answer:

$\lim_{x \to 0} [\frac{5^{x} + 3^{x} + 2^{x} - 3}{x}] = \lim_{x \to 0} [\frac{5^{x} - 1}{x} + \frac{3^{x} - 1}{x} + \frac{2^{x} - 1}{x}] = \log (5) + \log (3) + \log (2) = \log (5 \times 3 \times 2) = \log 30$

Page No 29.71:

Question 12:

$\lim_{x \to \infty} (a^{1 / x} - 1) x$

Answer:

$\lim_{x \to \infty} (a^{\frac{1}{x}} - 1) x Let y = \frac{1}{x} x \to \infty ∴ y \to 0 \Rightarrow \lim_{y \to 0} (\frac{a^{y} - 1}{y}) = \log a$

Page No 29.71:

Question 13:

$\lim_{x \to 0} \frac{a^{m x} - b^{n x}}{\sin k x}$

Answer:

$\lim_{x \to 0} [\frac{a^{m x} - b^{n x}}{\sin (k x)}] = \lim_{x \to 0} [\frac{(a^{m x} - 1) - (b^{n x} - 1)}{\sin k x}] Dividing the numerator and the denomiantor by x: = \lim_{x \to 0} [\frac{m (\frac{a^{m x} - 1}{m x}) - n (\frac{b^{n x} - 1}{n x})}{k \times \frac{\sin k x}{k x}}] = \frac{(m \log a - n \log b)}{k \times 1} = \frac{1}{k} [\log {(a)}^{m} - \log {(b)}^{n}] = \frac{1}{k} \log (\frac{a^{m}}{b^{n}})$

Page No 29.71:

Question 14:

$\lim_{x \to 0} \frac{a^{x} + b^{x} - c^{x} - d^{x}}{x}$

Answer:

$\lim_{x \to 0} [\frac{a^{n} + b^{n} - c^{n} - d^{n}}{x}] \lim_{x \to 0} [\frac{a^{n} + b^{n} - 2 - c^{n} - d^{n} + 2}{x}] = \lim_{x \to 0} [\frac{(a^{n} - 1) + (b^{n} - 1) - (c^{n} - 1) - (d^{n} - 1)}{x}] = \lim_{x \to 0} [(\frac{a^{n} - 1}{x}) + (\frac{b^{n} - 1}{x}) - (\frac{c^{n} - 1}{x}) - (\frac{d^{n} - 1}{x})] = \log a + \log b - \log c - \log d = (\log a + \log b) - (\log c + \log d) = \log (a b) - \log (c d) = \log (\frac{a b}{c d})$

Page No 29.71:

Question 15:

$\lim_{x \to 0} \frac{e^{x} - 1 + \sin x}{x}$

Answer:

$\lim_{x \to 0} [\frac{e^{x} - 1 + \sin x}{x}] = \lim_{x \to 0} [\frac{e^{x} - 1}{x} + \frac{\sin x}{x}] = 1 + 1 = 2$

Page No 29.71:

Question 16:

$\lim_{x \to 0} \frac{\sin 2 x}{e^{x} - 1}$

Answer:

$\lim_{x \to 0} [\frac{\sin (2 x)}{e^{x} - 1}]$

Dividing the numerator and the denominator by x:

$= \lim_{x \to 0} [\frac{\sin 2 x}{x \times \frac{e^{x} - 1}{x}}] = \lim_{x \to 0} [\frac{\sin 2 x}{2 x} \times \frac{2}{(\frac{e^{x} - 1}{x})}] = 1 \times \frac{2}{1} = 2$

Page No 29.71:

Question 17:

$\lim_{x \to 0} \frac{e^{\sin x} - 1}{x}$

Answer:

$\lim_{x \to 0} [\frac{e^{\sin x} - 1}{x}] = \lim_{x \to 0} [\frac{e^{\sin x} - 1}{\sin x} \times \frac{\sin x}{x}]$

x → 0
∴ sin x → 0

Let y=sin x

x → 0
∴ y → 0

$\Rightarrow \lim_{y \to 0} (\frac{e^{y} - 1}{y}) \times \lim_{x \to 0} (\frac{\sin x}{x}) = 1 \times 1$

Page No 29.71:

Question 18:

$\lim_{x \to 0} \frac{e^{2 x} - e^{x}}{\sin 2 x}$

Answer:

$\lim_{x \to 0} [\frac{e^{2 x} - e^{x}}{\sin (2 x)}] = \lim_{x \to 0} [\frac{e^{x} (e^{x} - 1)}{\sin 2 x}] = \lim_{x \to 0} [\frac{e^{x} (e^{x} - 1)}{x} \times \frac{2 x}{\sin 2 x} \times \frac{1}{2}] = e^{0} \times 1 \times \frac{1}{1} \times \frac{1}{2} = \frac{1}{2}$

Page No 29.71:

Question 19:

$\lim_{x \to a} \frac{\log x - \log a}{x - a}$

Answer:

$\lim_{x \to a} [\frac{\log x - \log a}{x - a}] = \lim_{x \to a} [\frac{\log (\frac{x}{a})}{a (\frac{x}{a} - 1)}] = \lim_{x \to a} [\frac{\log [1 + (\frac{x}{a} - 1)]}{a (\frac{x}{a} - 1)}] x \to a ∴ \frac{x}{a} \to 1 \Rightarrow \frac{x}{a} - 1 \to 0 Let y = \frac{x}{a} - 1 x \to a ∴ y \to 0 = \lim_{y \to 0} [\frac{\log (1 + y)}{a \times y}] = \frac{1}{a} \times 1 = \frac{1}{a}$

Page No 29.71:

Question 20:

$\lim_{x \to 0} \frac{\log (a + x) - \log (a - x)}{x}$

Answer:

$\lim_{x \to 0} [\frac{\log (a + x) - \log (a - x)}{x}] = \lim_{x \to 0} [\frac{\log (\frac{a + x}{a - x})}{x}] = \lim_{x \to 0} [\frac{\log (1 + \frac{a + x}{a - x} - 1)}{x}] = \lim_{x \to 0} [\frac{\log (1 + \frac{a + x - a + x}{a - x})}{x}] = \lim_{x \to 0} [\frac{\log (1 + \frac{2 x}{a - x})}{\frac{2 x}{a - x} \times (\frac{a - x}{2})}] x \to 0 ∴ \frac{2 x}{a - x} \to 0 Let y = \frac{2 x}{a - x} = \lim_{y \to 0} [\frac{\log (1 + y)}{y}] \times \lim_{x \to 0} (\frac{1}{\frac{a - x}{2}}) = 1 \times \frac{2}{a}$

Page No 29.71:

Question 21:

$\lim_{x \to 0} \frac{\log (2 + x) + \log 0.5}{x}$

Answer:

$\lim_{x \to 0} [\frac{\log (2 + x) + \log (0.5)}{x}] = \lim_{x \to 0} [\frac{\log ((2 + x) \times 0.5)}{x}] = \lim_{x \to 0} [\frac{\log (1 + \frac{x}{2})}{\frac{x}{2} \times 2}] = \frac{1}{2} \times 1 = \frac{1}{2}$

Page No 29.71:

Question 22:

$\lim_{x \to 0} \frac{\log (a + x) - \log a}{x}$

Answer:

$\lim_{x \to 0} [\frac{\log (a + x) - \log a}{x}] = \lim_{x \to 0} [\frac{\log (\frac{a + x}{a})}{x}] = \lim_{x \to 0} [\frac{\log (1 + \frac{x}{a})}{\frac{x}{a} \times a}] = \frac{1}{a} \times 1 = \frac{1}{a}$

Page No 29.71:

Question 23:

$\lim_{x \to 0} \frac{\log (3 + x) - \log (3 - x)}{x}$

Answer:

$\lim_{x \to 0} [\frac{\log (3 + x) - \log (3 - x)}{x}] = \lim_{x \to 0} [\frac{\log (\frac{3 + x}{3 - x})}{x}] = \lim_{x \to 0} [\frac{\log (1 + \frac{3 + x}{3 - x} - 1)}{x}] = \lim_{x \to 0} [\frac{\log (1 + \frac{3 + x - 3 + x}{3 - x})}{x}] = \lim_{x \to 0} [\frac{\log (1 + \frac{2 x}{3 - x})}{x}] = \lim_{x \to 0} [\frac{\log (1 + \frac{2 x}{3 - x})}{(\frac{2 x}{3 - x}) \times \frac{3 - x}{2}}] = \frac{1 \times 2}{3 - 0} = \frac{2}{3}$

Page No 29.71:

Question 24:

$\lim_{x \to 0} \frac{8^{x} - 2^{x}}{x}$

Answer:

$\lim_{x \to 0} [\frac{8^{x} - 2^{x}}{x}] = \lim_{x \to 0} [\frac{(8^{x} - 1)}{x} - \frac{(2^{x} - 1)}{x}] = \log (8) - \log (2) = \log (\frac{8}{2}) = \log 4$

Page No 29.71:

Question 25:

$\lim_{x \to 0} \frac{x (2^{x} - 1)}{1 - \cos x}$

Answer:

$\lim_{x \to 0} [\frac{x (2^{x} - 1)}{1 - \cos x}]$

Dividing the numerator and the denominator by x²:

$= \lim_{x \to 0} [(\frac{2^{x} - 1}{x}) \times \frac{x^{2}}{1 - \cos x}] = \lim_{x \to 0} [(\frac{2^{x} - 1}{x}) \times \frac{x^{2}}{2 \sin^{2} \frac{x}{2}}] = \lim_{x \to 0} [(\frac{2^{x} - 1}{x}) \times \frac{\frac{x^{2}}{4} \times 4}{2 \sin^{2} (\frac{x}{2})}] = (\log 2) \times \frac{4}{2} \times \frac{1}{1^{2}} = 2 \log 2 = \log {(2)}^{2} = \log 4$

Page No 29.71:

Question 26:

$\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\log (1 + x)}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 + x} - 1}{\log (1 + x)}]$

Rationalising the numerator:

$= \lim_{x \to 0} [\frac{(\sqrt{1 + x} - 1) (\sqrt{1 + x} + 1)}{\log (1 + x) (\sqrt{1 + x} + 1)}] = \lim_{x \to 0} [\frac{1 + x - 1}{\log (1 + x) (\sqrt{1 + x} + 1)}] = \frac{1}{1 \times (\sqrt{1 + 0} + 1)} = \frac{1}{2}$

Page No 29.71:

Question 27:

$\lim_{x \to 0} \frac{\log |1 + x^{3}|}{\sin^{3} x}$

Answer:

$\lim_{x \to 0} [\frac{\log (1 + x^{3})}{\sin^{3} x}] = \lim_{x \to 0} [\frac{\log (1 + x^{3})}{x^{3}} \times \frac{x^{3}}{\sin^{3} x}] = 1 \times \frac{1}{1^{3}} = 1$

Page No 29.71:

Question 28:

$\lim_{x \to π / 2} \frac{a^{\cot x} - a^{\cos x}}{\cot x - \cos x}$

Answer:

$\lim_{x \to \frac{π}{2}} [\frac{a^{\cot x} - a^{\cos x}}{\cot x - \cos x}] = \lim_{x \to \frac{π}{2}} [\frac{a^{\cos x} (a^{\cot x - \cos x} - 1)}{\cot x - \cos x}] x \to \frac{π}{2} ∴ \cot x - \cos x \to 0 \Rightarrow a^{\cos \frac{π}{2}} \times \log a = a^{0} \times \log a = \log a$

Page No 29.71:

Question 29:

$\lim_{x \to 0} \frac{e^{x} - 1}{\sqrt{1 - \cos x}}$

Answer:

$\lim_{x \to 0} [\frac{e^{x} - 1}{\sqrt{1 - \cos x}}] Rationalising the denominator, we get : = \lim_{x \to 0} [\frac{(e^{x} - 1)}{\sqrt{1 - \cos x}} \times \frac{\sqrt{1 + \cos x}}{\sqrt{1 + \cos x}}] = \lim_{x \to 0} [\frac{(e^{x} - 1) (\sqrt{1 + \cos x})}{\sqrt{1 - \cos^{2} x}}] = \lim_{x \to 0} [\frac{(e^{x} - 1) \sqrt{1 + \cos x}}{|\sin x|}]$

Dividing numerator and the denominator by x, we get:

$= \lim_{x \to 0} [(\frac{e^{x} - 1}{x}) \times \frac{\sqrt{1 + \cos x}}{(\frac{|\sin x|}{x})}] Left hand limit : \lim_{x \to 0^{-}} [(\frac{e^{x} - 1}{x}) \times \frac{\sqrt{1 + \cos x}}{(\frac{|\sin x|}{x})}] = \lim_{x \to 0^{-}} [(\frac{e^{x} - 1}{x}) \times \frac{\sqrt{1 + \cos x}}{(\frac{- \sin x}{x})}] = - \frac{1 \times \sqrt{2}}{1} = - \sqrt{2} Right hand limit : \lim_{x \to 0^{+}} [(\frac{e^{x} - 1}{x}) \times \frac{\sqrt{1 + \cos x}}{\frac{|\sin x|}{x}}] = \lim_{x \to 0^{+}} [(\frac{e^{x} - 1}{x}) \times \frac{\sqrt{1 + \cos x}}{\frac{\sin x}{x}}] = 1 \times \frac{\sqrt{2}}{1} = \sqrt{2}$

Left hand limit ≠ Right hand limit
Thus, limit does not exist.

Page No 29.71:

Question 30:

$\lim_{x \to 5} \frac{e^{x} - e^{5}}{x - 5}$

Answer:

$\lim_{x \to 5} [\frac{e^{x} - e^{5}}{x - 5}] = \lim_{x \to 5} e^{5} [\frac{e^{x - 5} - 1}{x - 5}] = e^{5} \times 1$

Page No 29.72:

Question 31:

$\lim_{x \to 0} \frac{e^{x + 2} - e^{2}}{x}$

Answer:

$\lim_{x \to 0} [\frac{e^{x + 2} - e^{2}}{x}] = \lim_{x \to 0} [e^{2} (\frac{e^{x} - 1}{x})] = e^{2} \times 1 = e^{2}$

Page No 29.72:

Question 32:

$\lim_{x \to π / 2} \frac{e^{\cos x} - 1}{\cos x}$

Answer:

$\lim_{x \to \frac{π}{2}} [\frac{e^{co s x} - 1}{\cos x}] If x \to \frac{π}{2}, then \cos x \to 0 . Let y = \cos x = \lim_{y \to 0} [\frac{e^{y} - 1}{y}] = 1$

Page No 29.72:

Question 33:

$\lim_{x \to 0} \frac{e^{3 + x} - \sin x - e^{3}}{x}$

Answer:

$\lim_{x \to 0} [\frac{e^{3 + x} - \sin x - e^{3}}{x}] = \lim_{x \to 0} [(\frac{e^{3 + x} - e^{3}}{x}) - \frac{\sin x}{x}] = \lim_{x \to 0} [e^{3} (\frac{e^{x} - 1}{x}) - \frac{\sin x}{x}] = e^{3} \times 1 - 1 = e^{3} - 1$

Page No 29.72:

Question 34:

$\lim_{x \to 0} \frac{e^{x} - x - 1}{2}$

Answer:

\lim_{x \to 0} [\frac{e^{x} - x - 1}{2}] e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . . . \infty = \lim_{x \to 0} [\frac{(1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} . . . . \infty) - x - 1}{2}] = \lim_{x \to 0} [\frac{\frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . . \infty}{2}] = 0

Page No 29.72:

Question 35:

$\lim_{x \to 0} \frac{e^{3 x} - e^{2 x}}{x}$

Answer:

$\lim_{x \to 0} [\frac{e^{3 x} - e^{2 x}}{x}] = \lim_{x \to 0} [(\frac{e^{3 x} - 1}{x}) - (\frac{e^{2 x} - 1}{x})] = \lim_{x \to 0} [3 (\frac{e^{3 x} - 1}{3 x}) - 2 (\frac{e^{2 x} - 1}{2 x})] = 3 \times 1 - 2 \times 1 = 1$

Page No 29.72:

Question 36:

$\lim_{x \to 0} \frac{e^{\tan x} - 1}{\tan x}$

Answer:

$\lim_{x \to 0} [\frac{e^{\tan x} - 1}{\tan x}]$

If x → 0, then tan x → 0.

Let y = tan x

$\underset{y \to 0}{= \lim} [\frac{e^{y} - 1}{y}] = 1$

Page No 29.72:

Question 37:

$\lim_{x \to 0} \frac{e^{b x} - e^{a x}}{x} where 0 < a < b$

Answer:

$\lim_{x \to 0} [\frac{e^{b x} - e^{a x}}{x}] = \lim_{x \to 0} [(\frac{e^{b x} - 1}{x}) - (\frac{e^{a x} - 1}{x})] = \lim_{x \to 0} [b (\frac{e^{b x} - 1}{b x}) - a \times (\frac{e^{a x} - 1}{a x})] = b \times 1 - a \times 1 = b - a$

Page No 29.72:

Question 38:

$\lim_{x \to 0} \frac{e^{\tan x} - 1}{x}$

Answer:

$\lim_{x \to 0} [\frac{e^{\tan x} - 1}{x}] = \lim_{x \to 0} [\frac{e^{\tan x} - 1}{\tan x} \times \frac{\tan x}{x}] = 1 \times 1$

Page No 29.72:

Question 39:

$\lim_{x \to 0} \frac{e^{x} - e^{\sin x}}{x - \sin x}$

Answer:

$\lim_{x \to 0} [\frac{e^{x} - e^{\sin x}}{x - \sin x}] = \lim_{x \to 0} [\frac{e^{\sin x} [\frac{e^{x}}{e^{\sin x}} - 1]}{x - \sin x}] = \lim_{x \to 0} e^{\sin x} [\frac{e^{x - \sin x} - 1}{x - \sin x}] = e^{\sin 0} = 1$

Page No 29.72:

Question 40:

$\lim_{x \to 0} \frac{3^{2 + x} - 9}{x}$

Answer:

$\lim_{x \to 0} [\frac{3^{2 + x} - 9}{x}] = \lim_{x \to 0} [\frac{3^{2} \cdot 3^{x} - 3^{2}}{x}] = 3^{2} \lim_{x \to 0} [\frac{3^{x} - 1}{x}] = 9 \log (3) = 9 \log_{e} 3$

Page No 29.72:

Question 41:

$\lim_{x \to 0} \frac{a^{x} - a^{- x}}{x}$

Answer:

$\lim_{x \to 0} [\frac{a^{x} - a^{- x}}{x}] = \lim_{x \to 0} [\frac{a^{x} - \frac{1}{a^{x}}}{x}] = \lim_{x \to 0} [\frac{a^{2 x} - 1}{a^{x} \cdot 2 x}] \times 2 = \lim_{x \to 0} [\frac{a^{2 x} - 1}{2 x}] \times \frac{2}{a^{x}} = \lim_{x \to 0} [\frac{a^{2 x} - 1}{2 x}] \times \frac{2}{a^{x}} = \log (a) \times \frac{2}{a^{0}} = 2 \log a$

Disclaimer: The answer given in the textbook is incorrect.

Page No 29.72:

Question 42:

$\lim_{x \to 0} \frac{x (e^{x} - 1)}{1 - \cos x}$

Answer:

$\lim_{x \to 0} [\frac{x (e^{x} - 1)}{1 - \cos x}]$

Dividing the numerator and the denominator by x²:

$= \lim_{x \to 0} [\frac{(\frac{e^{x} - 1}{x})}{(\frac{1 - \cos x}{x^{2}})}] = \lim_{x \to 0} [\frac{e^{x} - 1}{x} \times \frac{1}{\frac{2 \sin^{2} (\frac{x}{2})}{4 \times \frac{x^{2}}{4}}}] = 1 \times \frac{2}{1^{2}} = 2$

Page No 29.72:

Question 43:

$\lim_{x \to π / 2} \frac{2^{- \cos x} - 1}{x (x - \frac{π}{2})}$

Answer:

$\lim_{x \to \frac{π}{2}} [\frac{2^{- \cos x} - 1}{x (x - \frac{π}{2})}] = \lim_{x \to \frac{π}{2}} [\frac{2^{- \sin (\frac{π}{2} - x)} - 1}{x (x - \frac{π}{2})}] \{∵ \cos x = \sin (\frac{π}{2} - x)\} = \lim_{x \to \frac{π}{2}} [\frac{2^{\sin (x - \frac{π}{2})} - 1}{(x - \frac{π}{2}) \times x}] = \lim_{x \to \frac{π}{2}} [\frac{2^{\sin (x - \frac{π}{2})} - 1}{\sin (x - \frac{π}{2})} \times \frac{\sin (x - \frac{π}{2})}{(x - \frac{π}{2})} \times \frac{1}{x}] = \log_{e} 2 \times 1 \times \frac{1}{\frac{π}{2}} = \frac{2}{π} \log_{e} 2$

Page No 29.76:

Question 1:

$\lim_{n \to \infty} {(1 + \frac{x}{n})}^{n}$

Answer:

$\lim_{n \to \infty} {(1 + \frac{x}{n})}^{n} = e^{\lim_{n \to \infty} (\frac{x}{n}) \times n} = e^{x}$

Page No 29.76:

Question 2:

$\lim_{x \to 0^{+}} {\{1 + \tan^{2} \sqrt{x}\}}^{1 / 2 x}$

Answer:

$\lim_{x \to 0^{+}} [1 + \tan^{2} \sqrt{x}]^{\frac{1}{2 x}} Using the theorem given below : If \lim_{x \to a} f (x) = \lim_{x \to a} g (x) = 0 such that \lim_{x \to a} \frac{f (x)}{g (x)} exists, then \lim_{x \to a} {[1 + f (x)]}^{\frac{1}{g (x)}} = e^{\lim_{x \to a} \frac{f (x)}{g (x)}} . Here : f (x) = \tan^{2} \sqrt{x} g (x) = 2 x \Rightarrow e^{\lim_{x \to 0^{+}} (\frac{\tan^{2} \sqrt{x}}{2 x})} = e^{\lim_{x \to 0^{+}} (\frac{\tan \sqrt{x}}{\sqrt{x}}) \times (\frac{\tan \sqrt{x}}{\sqrt{x}}) \times \frac{1}{2}} = e^{1 \times 1 \times \frac{1}{2}} = \sqrt{e}$

Page No 29.76:

Question 3:

$\lim_{x \to 0} {(\cos x)}^{1 / \sin x}$

Answer:

$\lim_{x \to 0} {(\cos x)}^{\frac{1}{\sin x}} = \lim_{x \to 0} {[1 + \cos x - 1]}^{\frac{1}{\sin x}} Using the theorem given below: If \lim_{x \to a} f (x) = \lim_{x \to a} g (x) = 0 such that \lim_{x \to a} \frac{f (x)}{g (x)} exists, then \lim_{x \to a} {[1 + f (x)]}^{\frac{1}{g (x)}} = e^{\lim_{x \to a} \frac{f (x)}{g (x)}} . Here : f (x) = \cos x - 1 g (x) = \sin x \Rightarrow e^{\lim_{x \to 0} (\frac{\cos x - 1}{\sin x})} = e^{\lim_{x \to 0} (\frac{- 2 \sin^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}})} = e^{\lim_{x \to 0} (- \tan \frac{x}{2})} = e^{0} = 1$

Page No 29.76:

Question 4:

$\lim_{x \to 0} {(\cos x + \sin x)}^{1 / x}$

Answer:

$\lim_{x \to 0} {(\cos x + \sin x)}^{\frac{1}{x}} By adding and subtracting 1, we get : = \lim_{x \to 0} {[1 + \cos x + \sin x - 1]}^{\frac{1}{x}} Using the theorem given below : If \lim_{x \to a} f (x) = \lim_{x \to a} g (x) = 0 such that \lim_{x \to a} \frac{f (x)}{g (x)} exists, then \lim_{x \to a} {[1 + f (x)]}^{\frac{1}{g (x)}} = e^{\lim_{x \to a} \frac{f (x)}{g (x)}} . Here : f (x) = \cos x + \sin x - 1 g (x) = x \Rightarrow e^{\lim_{x \to 0}} (\frac{\cos x + \sin x - 1}{x}) \Rightarrow e^{\lim_{x \to 0}} [\frac{\sin x}{x} - \frac{(1 - \cos x)}{x}] \Rightarrow e^{\lim_{x \to 0}} (\frac{\sin x}{x} - \frac{2 \sin^{2} \frac{x}{2}}{x}) \Rightarrow e^{\lim_{x \to 0}} (\frac{\sin x}{x} - \frac{2 \sin (\frac{x}{2}) \times \sin (\frac{x}{2})}{2 \times \frac{x}{2}}) = e^{1 - 0} = e^{1}$

Page No 29.77:

Question 5:

$\lim_{x \to 0} {(\cos x + a \sin b x)}^{1 / x}$

Answer:

$\lim_{x \to 0} {(\cos x + a \sin b x)}^{\frac{1}{x}} = \lim_{x \to 0} {[1 + \cos x + a \sin b x - 1]}^{\frac{1}{x}} Using the theorem given below: If \lim_{x \to a} f (x) = \lim_{x \to a} g (x) = 0 such that \lim_{x \to a} \frac{f (x)}{g (x)} exists, then \lim_{x \to a} {[1 + f (x)]}^{\frac{1}{g (x)}} = e^{\lim_{x \to a} \frac{f (x)}{g (x)}} . Here : f (x) = \cos x + a \sin b x - 1 g (x) = x \Rightarrow e^{\lim_{x \to 0} [\frac{\cos x + a \sin b x - 1}{x}]} = e^{\lim_{x \to 0} [\frac{b \times a \sin b x}{b x} - \frac{(1 - \cos x)}{x}]} = e^{\lim_{x \to 0} (\frac{a b \sin b x}{b x} - \frac{2 \sin^{2} \frac{x}{2}}{x})} = e^{a b}$

Page No 29.77:

Question 6:

$\lim_{x \to \infty} {\{\frac{x^{2} + 2 x + 3}{2 x^{2} + x + 5}\}}^{\frac{3 x - 2}{3 x + 2}}$

Answer:

$\lim_{x \to \infty} {(\frac{x^{2} + 2 x + 3}{2 x^{2} + x + 5})}^{(\frac{3 x - 2}{3 x + 2})} = \lim_{x \to \infty} [1 + \frac{x^{2} + 2 x + 3}{2 x^{2} + x + 5} - 1] (^{\frac{3 x - 2}{3 x + 2}}) = \lim_{x \to \infty} {[1 + \frac{(x^{2} + 2 x + 3) - (2 x^{2} + x + 5)}{2 x^{2} + x + 5}]}^{(\frac{3 x - 2}{3 x + 2})} = \lim_{x \to \infty} {[1 + \frac{(- x^{2} + x - 2)}{2 x^{2} + x + 5}]}^{(\frac{3 x - 2}{3 x + 2})} = e^{\lim_{x \to \infty} (\frac{- x^{2} + x - 2}{2 x^{2} + x + 5}) \times (\frac{3 x - 2}{3 x + 2})} = e^{\lim_{x \to \infty} (\frac{- 1 + \frac{1}{x} - \frac{2}{x^{2}}}{2 + \frac{1}{x} + \frac{5}{x^{2}}}) \times (\frac{3 - \frac{2}{x}}{3 + \frac{2}{x}})} = e^{- \frac{1}{2} \times 1} = \frac{1}{\sqrt{e}}$

Disclaimer: The solution given in the book is incorrect. However, the solution given here has been created according to the question given in the book.

Page No 29.77:

Question 7:

$\lim_{x \to 1} {\{\frac{x^{3} + 2 x^{2} + x + 1}{x^{2} + 2 x + 3}\}}^{\frac{1 - \cos (x - 1)}{{(x - 1)}^{2}}}$

Answer:

$\lim_{x \to 1} {\{\frac{x^{3} + 2 x^{2} + x + 1}{x^{2} + 2 x + 3}\}}^{\frac{1 - \cos (x - 1)}{{(x - 1)}^{2}}} = \lim_{x \to 1} {\{1 + \frac{x^{3} + 2 x^{2} + x + 1}{x^{2} + 2 x + 3} - 1\}}^{\frac{1 - \cos (x - 1)}{{(x - 1)}^{2}}} = e^{\lim_{x \to 1} \{\frac{x^{3} + 2 x^{2} + x + 1 - x^{2} - 2 x - 3}{x^{2} + 2 x + 3}\} \times [\frac{1 - \cos (x - 1)}{{(x - 1)}^{2}}]} = e^{\lim_{x \to 1} \{\frac{x^{3} + x^{2} - x - 2}{x^{2} + 2 x + 3}\} \times [\frac{2 \sin^{2} \frac{(x - 1)}{2}}{4 \times \frac{{(x - 1)}^{2}}{4}}]} = e^{(\frac{1 + 1 - 1 - 2}{1 + 2 + 3}) \times \frac{1}{2}} = e^{- \frac{1}{12}}$

Disclaimer: The solution given in the book is incorrect. However, the solution given here has been created according to the question given in the book.

Page No 29.77:

Question 8:

$\lim_{x \to 0} {\{\frac{e^{x} + e^{- x} - 2}{x^{2}}\}}^{1 / x^{2}}$

Answer:

$\lim_{x \to 0} {[\frac{e^{x} + e^{- x} - 2}{x^{2}}]}^{(\frac{1}{x^{2}})} = \lim_{x \to 0} {[1 + \frac{e^{x} + e^{- x} - 2}{x^{2}} - 1]}^{(\frac{1}{x^{2}})} = e^{\lim_{x \to 0} (\frac{e^{x} + e^{- x} - 2}{x^{2}} - 1) \times (\frac{1}{x^{2}})} ∵ e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . . . . \propto e^{- x} = 1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + . . . . . . \propto \Rightarrow e^{x} + e^{- x} = 2 + \frac{2 x^{2}}{2!} + \frac{2 x^{4}}{4!} + . . . . . \propto = e^{\lim_{x \to 0} (\frac{2 + \frac{2 x^{2}}{2!} + \frac{2 x^{4}}{4!} . . . \propto - 2}{x^{2}} - 1) \times (\frac{1}{x^{2}})} = e^{\lim_{x \to 0} (\frac{\frac{2 x^{2}}{2!} + \frac{2 x^{4}}{4!} + . . . . . . \propto}{x^{4}} - \frac{1}{x^{2}})} = e^{\lim_{x \to 0} (\frac{x^{2} + \frac{x^{4}}{12} + . . . . . \propto - x^{2}}{x^{4}})} = e^{\frac{1}{12}}$

Page No 29.77:

Question 9:

$\lim_{x \to a} {\{\frac{\sin x}{\sin a}\}}^{\frac{1}{x - a}}$

Answer:

$\lim_{x \to a} {\{\frac{\sin x}{\sin a}\}}^{\frac{1}{(x - a)}} = \lim_{x \to a} {\{1 + \frac{\sin x}{\sin a} - 1\}}^{\frac{1}{(x - a)}} = \lim_{x \to a} {\{1 + \frac{\sin x - \sin a}{\sin a}\}}^{\frac{1}{(x - a)}} = e^{\lim_{x \to a} (\frac{\sin x - \sin a}{\sin a}) \times \frac{1}{(x - a)}} = e^{\lim_{x \to a} (\frac{2 \cos (\frac{x + a}{2}) \sin (\frac{x - a}{2})}{\sin a \times (\frac{x - a}{2}) \times 2})} = e^{\frac{\cos a}{\sin a}} = e^{\cot a}$

Page No 29.77:

Question 10:

$\lim_{x \to \infty} {\{\frac{3 x^{2} + 1}{4 x^{2} - 1}\}}^{\frac{x^{3}}{1 + x}}$

Answer:

$\lim_{x \to \infty} {\{\frac{3 x^{2} + 1}{4 x^{2} - 1}\}}^{\frac{x^{3}}{1 + x}} = \lim_{x \to \infty} {\{1 + \frac{3 x^{2} + 1}{4 x^{2} - 1} - 1\}}^{\frac{x^{3}}{1 + x}} = \lim_{x \to \infty} {\{1 + \frac{3 x^{2} + 1 - 4 x^{2} + 1}{4 x^{2} - 1}\}}^{\frac{x^{3}}{1 + x}} = \lim_{x \to \infty} {\{1 + \frac{2 - x^{2}}{4 x^{2} - 1}\}}^{\frac{x^{3}}{1 + x}} = e^{\lim_{x \to \infty} (\frac{2 - x^{2}}{4 x^{2} - 1}) \times (\frac{x^{3}}{1 + x})} = e^{\lim_{x \to \infty} (\frac{- x^{5} + 2 x^{3}}{(4 x^{2} - 1) (1 + x)})} Dividing N^{r} and D^{r} by x^{5} : = e^{\lim_{x \to \infty} (\frac{- 1 + \frac{2}{x^{3}}}{\frac{(4 x^{2} - 1)}{x^{3}} \frac{(1 + x)}{x^{2}}})} = e^{\lim_{x \to \infty} (\frac{- 1 + \frac{2}{x^{3}}}{(\frac{4}{x} - \frac{1}{x^{3}}) (\frac{1}{x^{2}} + \frac{1}{x})})} = e^{- \frac{1}{0}} = e^{- \infty} = 0$

Page No 29.77:

Question 1:

$\lim_{n \to \infty} \frac{1^{2} + 2^{2} + 3^{2} + . . . + n^{2}}{n^{3}}$ is equal to

(a) 1
(b) 1/2
(c) 1/3
(d) 0

Answer:

(c) 1/3

$\lim_{n \to \infty} \frac{1^{2} + 2^{2} + 3^{2} . . . . . n^{2}}{n^{3}} = \lim_{n \to \infty} \frac{Σ n^{2}}{n^{3}} = \lim_{n \to \infty} \frac{n (n + 1) (2 n + 1)}{6 n^{3}} = \lim_{n \to 0} \frac{(n + 1) (2 n + 1)}{6 n^{2}} {Dividing the numerator and the denominator by n}^{2}, we get: \lim_{n \to \infty} \frac{\frac{(n + 1)}{n} \times \frac{(2 n + 1)}{n}}{6} = \lim_{n \to \infty} \frac{(1 + \frac{1}{n}) (2 + \frac{1}{n})}{6} \Rightarrow \frac{2}{6} = \frac{1}{3}$

Page No 29.77:

Question 2:

$\lim_{x \to 0} \frac{\sin 2 x}{x}$ is equal to

(a) 0
(b) 1
(c) 1/2
(d) 2

Answer:

(d) 2
$\lim_{x \to 0} \frac{\sin 2 x}{x} = \lim_{x \to 0} 2 (\frac{\sin 2 x}{2 x}) = 2 \times 1 = 2$

Page No 29.77:

Question 3:

If $f (x) = x \sin (1 / x), x \neq 0,$ then $\lim_{x \to 0} f (x) =$
(a) 1
(b) 0
(c) −1
(d) does not exist

Answer:

(b) 0
$\lim_{x \to 0} f (x) = \lim_{x \to 0} x \sin \frac{1}{x}, x \neq 0 LHL at x = 0 : \lim_{x \to 0^{-}} f (x) = \lim_{h \to 0} f (0 - h) \Rightarrow \lim_{h \to 0} - h \sin (\frac{1}{- h}) = 0 RHL at x = 0 : \lim_{x \to 0^{+}} f (x) = \lim_{h \to 0} f (0 + h) \Rightarrow \lim_{h \to 0} h \sin (\frac{1}{h}) = 0 \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x) Hence, \lim_{x \to 0} f (x) exists and is equal to 0 .$

Page No 29.77:

Question 4:

$\lim_{x \to 0} \frac{1 - \cos 2 x}{x} is$

(a) 0
(b) 1
(c) 2
(d) 4

Answer:

(a) 0

$\lim_{x \to 0} \frac{1 - \cos 2 x}{x} = \lim_{x \to 0} \frac{2 \sin^{2} x}{x} = \lim_{x \to 0} 2 x \times \frac{\sin^{2} x}{x^{2}} \Rightarrow 0$

Page No 29.77:

Question 5:

$\lim_{x \to 0} \frac{(1 - \cos 2 x) \sin 5 x}{x^{2} \sin 3 x} =$

(a) 10/3
(b) 3/10
(c) 6/5
(d) 5/6

Answer:

(a) 10/3

$\lim_{x \to 0} \frac{(1 - \cos 2 x) \sin 5 x}{x^{2} \sin 3 x} = \lim_{x \to 0} \frac{2 \sin^{2} x \sin 5 x}{x^{2} \sin 3 x} = \lim_{x \to 0} \frac{\frac{2 \sin^{2} x}{x^{2}} (\frac{\sin 5 x}{5 x}) \times 5}{(\frac{\sin 3 x}{3 x}) 3} = \frac{10 \times 1^{2} \times 1}{1 \times 3} = \frac{10}{3}$

Page No 29.77:

Question 6:

$\lim_{x \to 0} \frac{x}{\tan x} is$

(a) 0
(b) 1
(c) 4
(d) not defined

Answer:

(b) 1

$\lim_{x \to 0} \frac{x}{\tan x} = \lim_{x \to 0} \frac{1}{\frac{\tan x}{x}} = \frac{1}{1} = 1$

Page No 29.77:

Question 7:

$\lim_{n \to \infty} \{\frac{1}{1 - n^{2}} + \frac{2}{1 - n^{2}} + . . . + \frac{n}{1 - n^{2}}\}$ is equal to

(a) 0
(b) −1/2
(c) 1/2
(d) none of these

Answer:

(b) $\frac{- 1}{2}$

$\lim_{n \to \infty} [\frac{1}{1 - n^{2}} + \frac{2}{1 - n^{2}} + . . . . . . + \frac{n}{1 - n^{2}}] = \lim_{n \to \infty} (\frac{1}{1 - n^{2}}) [1 + 2 + 3 . . . . . n] = \lim_{n \to \infty} (\frac{1}{1 - n^{2}}) [\frac{n (n + 1)}{2}] = \lim_{n \to \infty} \frac{1}{2} [\frac{n (n + 1)}{(1 - n) (1 + n)}] = \lim_{n \to \infty} \frac{1}{2} \frac{n}{1 - n} = \lim_{n \to \infty} \frac{1}{2} \frac{1}{\frac{1}{n} - 1} = \frac{1}{2} (- 1) = \frac{- 1}{2}$

Page No 29.77:

Question 8:

$\lim_{x \to \infty} \frac{\sin x}{x}$ equals

(a) 0
(b) ∞
(c) 1
(d) does not exist

Answer:

(a)
$\lim_{x \to \infty} \frac{\sin x}{x} L et x = \frac{1}{y} x \to \infty ∴ y \to 0 = \lim_{y \to 0} \frac{\sin \frac{1}{y}}{\frac{1}{y}} = \lim_{y \to 0} y \sin \frac{1}{y} = \lim_{y \to 0} y \times \lim_{y \to 0} \sin \frac{1}{y} = 0 \times \lim_{y \to 0} \sin \frac{1}{y} = 0$

Page No 29.77:

Question 9:

$\lim_{x \to 0} \frac{\sin x^{0}}{x}$ is equal to

(a) 1
(b) π
(c) x
(d) π/180

Answer:

(d) $\frac{π}{180}$

$\lim_{x \to 0} \frac{\sin x^{0}}{x} = \lim_{x \to 0} \frac{\sin \frac{π}{180} x}{x} = \lim_{x \to 0} \frac{\sin (\frac{π}{180} x)}{(\frac{π}{180} x)} \times \frac{π}{180} = \frac{π}{180} \times 1 = \frac{π}{180}$

Page No 29.78:

Question 10:

$\lim_{x \to 3} \frac{x - 3}{|x - 3|},$ is equal to

(a) 1
(b) −1
(c) 0
(d) does not exist

Answer:

(d) does not exist

$\lim_{x \to 3} \frac{x - 3}{|x - 3|} LHL at x = 3 : \lim_{x \to 3^{-}} \frac{x - 3}{- (x - 3)} [∵ |x - 3| = - (x - 3), when x < 3] \Rightarrow - 1 RHL at x = 3 : \lim_{x \to 3^{+}} \frac{x - 3}{x - 3} [∵ |x - 3| = x - 3, when x > 3] = 1$

LHL ≠ RHL
Therefore, limit does not exist.

Page No 29.78:

Question 11:

$\lim_{x \to a} \frac{x^{n} - a^{n}}{x - a}$ is equal at

(a) naⁿ
(b) naⁿ⁻¹
(c) na
(d) 1

Answer:

(b) naⁿ⁻¹

$\lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = \lim_{x \to a^{+}} \frac{x^{n} - a^{n}}{x - a} [∵ f (x) exists, \lim_{x \to a} f (x) = \lim_{x \to a^{+}} f (x)] = \lim_{h \to 0} \frac{{(a + h)}^{n} - a^{n}}{a + h - a} = \lim_{h \to 0} a^{n} \frac{[{(1 + \frac{h}{a})}^{n} - 1]}{h} = a^{n} \lim_{h \to 0} [1 + n \cdot \frac{h}{a} + \frac{n (n - 1)}{2!} \frac{h^{2}}{a^{2}} . . . + . . . - 1] = a^{n} \lim_{h \to 0} [\frac{n}{a} + \frac{h (h - 1)}{2!} \frac{h}{a^{2}} + . . .] = a^{n} \frac{n}{a} = n a^{n - 1}$

Page No 29.78:

Question 12:

$\lim_{x \to π / 4} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}$ is equal to

(a) $\frac{1}{\sqrt{2}}$

(b) $\frac{1}{2}$

(c) $\frac{1}{2 \sqrt{2}}$

(d) 1

Answer:

(b) 1/2

$\lim_{x \to \frac{π}{4}} \frac{\sqrt{2} \cos x - 1}{\cot x - 1} Rationalising the numerator, we get : = \lim_{x \to \frac{π}{4}} (\frac{\sqrt{2} \cos x - 1}{\cot x - 1}) \times (\frac{\sqrt{2} \cos x + 1}{\sqrt{2} \cos x + 1}) = \lim_{x \to \frac{π}{4}} \frac{(2 \cos^{2} x - 1)}{(\cos x - \sin x)} \times \frac{\sin x}{(\sqrt{2} \cos x + 1)} = \lim_{x \to \frac{π}{4}} \frac{(\cos^{2} x - \sin^{2} x)}{(\cos x - \sin x)} \times \frac{\sin x}{(\sqrt{2} \cos x + 1)} = \lim_{x \to \frac{π}{4}} \frac{(\cos x + \sin x) \sin x}{(\sqrt{2} \cos x + 1)} = \frac{(\cos \frac{π}{4} + \sin \frac{π}{4}) \sin \frac{π}{4}}{(\sqrt{2} \cos \frac{π}{4} + 1)} = \frac{(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) (\frac{1}{\sqrt{2}})}{\sqrt{2} . \frac{1}{\sqrt{2}} + 1} = \frac{(\frac{2}{\sqrt{2}}) \times \frac{1}{\sqrt{2}}}{2} = \frac{1}{2}$
The correct answer is B.

Page No 29.78:

Question 13:

$\lim_{x \to \infty} \frac{\sqrt{x^{2} - 1}}{2 x + 1}$ is equal to

(a) 1
(b) 0
(c) −1
(d) 1/2

Answer:

(d) 1/2

$\lim_{x \to \infty} \frac{\sqrt{x^{2} - 1}}{2 x + 1} = \lim_{x \to \infty} \frac{\sqrt{1 - \frac{1}{x^{2}}}}{2 + \frac{1}{x}} Divid ing the numerator and the denominator by x, we get : = \frac{\sqrt{1 - 0}}{2 + 0} = \frac{1}{2}$

Hence, the correct answer is d.

Page No 29.78:

Question 14:

$\underset{h \to 0}{\lim 2} \{\frac{\sqrt{3} \sin (π / 6 + h) - \cos (π / 6 + h)}{\sqrt{3} h (\sqrt{3} \cos h - \sin h)}\}$ is equal to

(a) 2/3
(b) 4/3
(c) $- 2 \sqrt{3}$
(d) −4/3

Answer:

(d) 4/3

$\lim_{h \to 0} 2 [\frac{\sqrt{3} \sin (π / 6 + h) - \cos (π / 6 + h)}{\sqrt{3} h (\sqrt{3} \cos h - \sin h)}] = \lim_{h \to 0} 2 \frac{[\frac{\sqrt{3}}{2} \cos h + \frac{3}{2} \sin h - \frac{\sqrt{3}}{2} \cos h + \frac{\sin h}{2}]}{h (3 \cos h - \sqrt{3} \sin h)} = \lim_{h \to 0} 2 (\frac{2 \sin h}{h}) \times \frac{1}{(3 \cos h - \sqrt{3} \sin h)} = \lim_{h \to 0} \frac{4}{3 \cos h - \sqrt{3} \sin h} = \frac{4}{3}$

Page No 29.78:

Question 15:

$\lim_{h \to 0} \{\frac{1}{h \sqrt[3]{8 + h}} - \frac{1}{2 h}\} =$

(a) −1/12
(b) −4/3
(c) −16/3
(d) −1/48

Answer:

(d) −1/48

$\lim_{h \to 0} [\frac{1}{h \sqrt[3]{8 + h}} - \frac{1}{2 h}] = \lim_{h \to 0} [\frac{1}{h} \{\frac{1}{\sqrt[3]{8 + h}} - \frac{1}{2}\}] = \lim_{h \to 0} [\frac{1}{h} \{\frac{2 - {(8 + h)}^{1 / 3}}{2 \times \sqrt[3]{8 + h}}\}] = \lim_{h \to 0} [\frac{1}{h} \{\frac{8^{1 / 3} - {(8 + h)}^{1 / 3}}{2 \sqrt[3]{8 + h}}\}] [A^{3} - B^{3} = (A - B) (A^{2} + A B + B^{2}) o r A - B = \frac{A^{3} - B^{3}}{A^{2} + A B + B^{2}}] = \lim_{h \to 0} [\frac{8 - (8 + h)}{h \{2 \sqrt[3]{8 + h}\} \{4 + 2 {(8 + h)}^{1 / 3} + {(8 + h)}^{2 / 3}\}}] = [\frac{- 1}{2 \times \sqrt[3]{8} \{4 + 2 \times 8^{1 / 3} + 8^{2 / 3}\}}] = \frac{- 1}{2 \times 2 (4 + 4 + 4)} = \frac{- 1}{48}$

Page No 29.78:

Question 16:

$\lim_{n \to \infty} \{\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + . . . + \frac{1}{(2 n + 1) (2 n + 3)}\}$ is equal to

(a) 0
(b) 1/2
(c) 1/9
(d) 2

Answer:

(b) 1/2

$\lim_{n \to \infty} [\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} . . . \frac{1}{(2 n + 1) (2 n + 3)}] Here, T_{n} = \frac{1}{(2 n - 1) (2 n + 1)} \Rightarrow T_{n} = \frac{A}{(2 n - 1)} + \frac{B}{(2 n + 1)} On equating A = \frac{1}{2} and B = - \frac{1}{2} : T_{n} = \frac{1}{2 (2 n - 1)} - \frac{1}{2 (2 n + 1)} \Rightarrow T_{1} = \frac{1}{2} [1 - \frac{1}{3}] \Rightarrow T_{2} = \frac{1}{2} [\frac{1}{3} - \frac{1}{5}] \Rightarrow T_{n - 1} = \frac{1}{2} [\frac{1}{2 n - 1} - \frac{1}{2 n - 1}] \Rightarrow T_{n} = \frac{1}{2} [\frac{1}{2 n - 1} - \frac{1}{2 n + 1}] \Rightarrow T_{1} + T_{2} + T_{3} . . . T_{n} = \frac{1}{2} [1 - \frac{1}{2 n + 1}] \Rightarrow T_{1} + T_{2} + T_{3} . . . T_{n} = \frac{1}{2} [\frac{2 n}{2 n + 1}] \Rightarrow T_{1} + T_{2} + T_{3} . . . T_{n} = \frac{n}{2 n + 1} ∴ \lim_{n \to \infty} [\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} . . . \frac{1}{(2 n + 1) (2 n + 3)}] = \lim_{n \to \infty} [\sum_{n = 1}^{n} \frac{1}{(2 n - 1) (2 n + 1)}] = \lim_{n \to \infty} (\frac{n}{2 n + 1})$

$= \lim_{n \to \infty} (\frac{1}{2 + \frac{1}{n}}) [Divid ing N^{r} and D^{r} by n] = \frac{1}{2}$

Page No 29.78:

Question 17:

$\lim_{x \to 1} \frac{\sin π x}{x - 1}$ is equal to

(a) −π

(b) π

(c) $- \frac{1}{π}$

(d) $\frac{1}{π}$

Answer:

(a) −π
$\lim_{x \to 1} \frac{\sin π x}{x - 1} = \lim_{h \to 0} \frac{\sin π (1 + h)}{(1 + h) - 1} = \lim_{h \to 0} \frac{\sin (π + π h)}{h} = \lim_{h \to 0} \frac{- \sin π h}{h} = \lim_{h \to 0} - (\frac{\sin π h}{π h}) π = - π$

Page No 29.78:

Question 18:

If $\lim_{x \to 1} \frac{x + x^{2} + x^{3} + . . . + x^{n} - n}{x - 1} = 5050$ then n equal

(a) 10
(b) 100
(c) 150
(d) none of these

Answer:

(b) 100

$\lim_{x \to 1} \frac{x + x^{2} + x^{3} . . . x^{n} - n}{x - 1} = 5050 \Rightarrow \lim_{x \to 1} \frac{(x - 1)}{x - 1} + \frac{(x^{2} - 1)}{x - 1} + \frac{(x^{3} - 1)}{x - 1} . . . \frac{x^{n} - 1}{x - 2} = 5050 \Rightarrow 1 + 2 + 3 . . . n = 5050 [∵ \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}] \Rightarrow \frac{n (n + 1)}{2} = 5050 \Rightarrow n (n + 1) = 10100 \Rightarrow n (n + 1) = 100 \times 101 On c omparing : n = 100$

Page No 29.78:

Question 19:

The value of $\lim_{x \to \infty} \frac{\sqrt{1 + x^{4}} + (1 + x^{2})}{x^{2}}$ is

(a) −1
(b) 1
(c) 2
(d) none of these

Answer:

(c) 2

$\lim_{x \to \infty} \frac{\sqrt{1 + x^{4}} + (1 + x^{2})}{x^{2}} = \lim_{x \to \infty} \sqrt{\frac{1}{x^{4}} + 1} + \frac{1}{x^{2}} + 1 = 2$

Page No 29.78:

Question 20:

$\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}$ is equal to

(a) $\frac{1}{2}$

(b) 2

(c) 0

(d) 1

Answer:

(a) $\frac{1}{2}$

$\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x} = \lim_{x \to 0} \frac{(\sqrt{1 + x} - 1) (\sqrt{1 + x} + 1)}{(\sqrt{1 + x} + 1) x} = \lim_{x \to 0} \frac{1 + x - 1}{x \sqrt{1 + x + 1}} = \lim_{x \to 0} \frac{1}{x (\sqrt{1 + x} + 1)} = \frac{1}{2}$

Page No 29.79:

Question 21:

$\lim_{x \to π / 3} \frac{\sin (\frac{π}{3} - x)}{2 \cos x - 1}$ is equal to

(a) $\sqrt{3}$

(b) $\frac{1}{2}$

(c) $\frac{1}{\sqrt{3}}$

(d) $\sqrt{3}$

Answer:

(c) $\frac{1}{\sqrt{3}}$

$\lim_{x \to \frac{π}{3}} \frac{\sin (\frac{π}{3} - x)}{2 \cos x - 1} = \lim_{h \to 0} \frac{\sin \frac{π}{3} - (\frac{π}{3} - h) []}{2 \cos (\frac{π}{3} - h) - 1} = \lim_{h \to 0} \frac{\sin h}{2 [\cos \frac{π}{3} \cos h + \sin \frac{π}{3} \sin h] - 1} = \lim_{h \to 0} \frac{\sin h}{2 [\frac{1}{2} \cos h + \frac{\sqrt{3}}{2} \sin h] - 1} = \lim_{h \to 0} \frac{\sin h}{\cos h + \sqrt{3} \sin h - 1} = \lim_{h \to 0} \frac{\sin h}{- 2 \sin^{2} \frac{h}{2} + \sqrt{3} \sin h} Dividing N^{r} and D^{r} by h : = \lim_{h \to 0} \frac{\frac{\sin h}{h}}{- (2 \times \frac{h}{4}) (\frac{\sin^{2} \frac{h}{2}}{h \times \frac{h}{4}}) + \frac{\sqrt{3} \sin h}{h}} = \frac{1}{\sqrt{3}}$

Page No 29.79:

Question 22:

$\lim_{x \to 3} \frac{\sum_{r = 1}^{n} x^{r} - \sum_{r = 1}^{n} 3^{r}}{x - 3}$ is real to

(a) $\frac{(2 n - 1) \times 3^{n}}{4}$

(b) $\frac{(2 n - 1) \times 3^{n} + 1}{4}$

(c) $(2 n - 1) 3^{n} + 1$

(d) $\frac{(2 n - 1) \times 3^{n} - 1}{4}$

Answer:

(b) $\frac{(2 n - 1) \times 3^{n} + 1}{4}$

$\lim_{x \to 3} \frac{\sum_{r = 1}^{n} x^{r} - \sum_{r = 1}^{n} 3^{r}}{x - 3} = \lim_{x \to 3} \frac{x^{1} + x^{2} + x^{3} + . . . . . . + x^{n} - (3^{1} + 3^{2} + 3^{3} . . . . . 3^{n})}{x - 3} = \lim_{x \to 3} \frac{x - 3}{x - 3} + \frac{x^{2} - 3^{2}}{x - 3} + \frac{x^{3} - 3^{3}}{x - 3} + . . . . \frac{x^{n} - 3^{n}}{x - 3} = 1 + 2 \times 3 + 3 \times 3^{2} + . . . . . + n 3^{n - 1} [∵ \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}]$

This series is an AGP of the form given below:

S = 1 + 2r + 3r²........nrⁿ^–1
$S = \frac{1 - r^{n}}{{(1 - r)}^{2}} - \frac{n r^{n}}{1 - r} r = 3, a = 1 d = 1 = \frac{1 - 3^{n} + 2 n 3^{n}}{4} = \frac{3^{n} (2 n - 1) + 1}{4}$

Page No 29.79:

Question 23:

$\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + (2 n - 1) - 2 n}{\sqrt{n^{2} + 1} + \sqrt{n^{2} - 1}}$ is equal to

(a) $\frac{1}{2}$

(b) $- \frac{1}{2}$

(c) 1

(d) −1

Answer:

(b) $- \frac{1}{2}$

$\lim_{n \to \infty} [\frac{1 - 2 + 3 - 4 + 5 - 6 + . . . (2 n - 1) - 2 n}{\sqrt{n^{2} + 1} + \sqrt{n^{2} - 1}}] = \lim_{n \to \infty} [\frac{(1 + 3 + 5 + . . . 2 n - 1) - (2 + 4 + 6 + . . . 2 n)}{(\sqrt{n^{2} + 1} + \sqrt{n^{2} - 1})}] = \lim_{n \to \infty} [\frac{\frac{n}{2} (1 + 2 n - 1) - \frac{n}{2} (2 + 2 n)}{(\sqrt{n^{2} + 1} + \sqrt{n^{2} - 1})}] = \lim_{n \to \infty} [\frac{n^{2} - n (n + 1)}{(\sqrt{n^{2} + 1} + \sqrt{n^{2} - 1})}] = \lim_{n \to \infty} [\frac{- n}{(\sqrt{n^{2} + 1} + \sqrt{n^{2} - 1})}]$

Dividing the numerator and the denominator by n:

$= \lim_{n \to \infty} [\frac{- 1}{\sqrt{1 + \frac{1}{n^{2}}} + \sqrt{1 - \frac{1}{n^{2}}}}] = \frac{- 1}{1 + 1} = \frac{- 1}{2}$

Page No 29.79:

Question 24:

If $f (x) = \{\begin{array}{l} x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{array},$ then $\lim_{x \to 0} f (x)$ equals

(a) 1
(b) 0
(c) −1
(d) none of these

Answer:

(b) 0

$f (x) = \{\begin{cases} x \sin (\frac{1}{x}), x \neq 0 \\ 0, x = 0 \end{cases}$

LHL:

$\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} [x \sin (\frac{1}{x})]$

Let x = 0 – h, where h → 0.

$= \lim_{h \to 0} [(- h) \times \sin (- \frac{1}{h})]$

= 0 × The oscillating number between –1 and 1
= 0
RHL
$\lim_{x \to 0^{+}} f (x)$

Let x = 0 + h, where h → 0.

$= \lim_{h \to 0} [h \times \sin (\frac{1}{h})]$

= 0 × The oscillating number between –1 and 1
= 0
LHL = RHL = 0
$∴ \lim_{x \to 0} f (x) = 0$

Page No 29.79:

Question 25:

$\lim_{n \to \infty} \frac{n!}{(n + 1)! + n!}$ is equal to

(a) $\frac{1}{2}$

(b) 0

(c) 2

(d) 1

Answer:

(b) 0

$\lim_{n \to \infty} [\frac{n!}{(n + 1)! + n!}] = \lim_{n \to \infty} [\frac{n!}{(n + 1) \times n! + n!}] = \lim_{n \to \infty} [\frac{n!}{n! (n + 1 + 1)}] = \lim_{n \to \infty} [\frac{1}{n + 2}] = 0$

Page No 29.79:

Question 26:

$\lim_{x \to π / 4} \frac{4 \sqrt{2} - {(\cos x + \sin x)}^{5}}{1 - \sin 2 x}$ is equal to

(a) $5 \sqrt{2}$

(b) $3 \sqrt{2}$

(c) $\sqrt{2}$

(d) none of these

Answer:

(a)

$\lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} - {(\cos x + \sin x)}^{5}}{1 - \sin 2 x} = \lim_{x \to \frac{π}{4}} \frac{2^{\frac{5}{2}} - {({(\cos x + \sin x)}^{2})}^{\frac{5}{2}}}{2 - (1 + \sin 2 x)} = \lim_{x \to \frac{π}{4}} \frac{2^{\frac{5}{2}} - {({(\cos x + \sin x)}^{2})}^{\frac{5}{2}}}{2 - {(\cos x + \sin x)}^{2}} Let t = {(\cos x + \sin x)}^{2} x \to \frac{π}{4} ∴ t = {(\cos \frac{π}{4} + \sin \frac{π}{4})}^{2} \to {(\sqrt{2})}^{2} = 2 = \lim_{t \to 2} \frac{2^{\frac{5}{2}} - {(t)}^{\frac{5}{2}}}{2 - (t)} = \frac{5}{2} {(2)}^{\frac{3}{2}} [∵ \lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}] = 5 \sqrt{2}$

Page No 29.79:

Question 27:

$\lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x} - \sqrt{3}}}{x - 2}$ is equal to

(a) $\frac{1}{8 \sqrt{3}}$

(b) $\frac{1}{\sqrt{3}}$

(c) $8 \sqrt{3}$

(d) $\sqrt{3}$

Answer:

$(a) ∵ \lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x}} - \sqrt{3}}{x - 2} = \lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x}} - \sqrt{3}}{x - 2} \times \frac{\sqrt{1 + \sqrt{2 + x}} + \sqrt{3}}{\sqrt{1 + \sqrt{2 + x}} + \sqrt{3}} = \lim_{x \to 2} \frac{\sqrt{2 + x} - 2}{(x - 2) (\sqrt{1 + \sqrt{2 + x}} + \sqrt{3})} = \lim_{x \to 2} \frac{\sqrt{2 + x} - 2}{(x - 2) (\sqrt{1 + \sqrt{2 + x}} + \sqrt{3})} \times \frac{\sqrt{2 + x} + 2}{\sqrt{2 + x} + 2} = \lim_{x \to 2} \frac{(x - 2)}{(x - 2) (\sqrt{1 + \sqrt{2 + x}} + \sqrt{3}) (\sqrt{2 + x} + 2)} = \lim_{x \to 2} \frac{1}{(\sqrt{1 + \sqrt{2 + x}} + \sqrt{3}) (\sqrt{2 + x} + 2)} = \frac{1}{(\sqrt{1 + \sqrt{2 + 2}} + \sqrt{3}) (\sqrt{2 + 2} + 2)} = \frac{1}{2 \sqrt{3} \times 4} = \frac{1}{8 \sqrt{3}}$

Disclaimer: The question in the book has some error, which has been resolved here and the solution is created accordingly.

Page No 29.79:

Question 28:

$\lim_{x \to \infty} a^{x} \sin (\frac{b}{a^{x}}), a, b > 1$ is equal to

(a) b
(b) a
(c) a log_e b
(d) b log_e a

Answer:

(a) b

$\lim_{x \to \infty} a^{x} \sin (\frac{b}{a^{x}}) \lim_{x \to \infty} b (\frac{\sin \frac{b}{a^{x}}}{\frac{b}{a^{x}}}) Let \frac{b}{a^{x}} = y x \to \infty ∴ y \to 0 \lim_{y \to 0} \frac{b \sin y}{y} = b$

Page No 29.79:

Question 29:

$\lim_{x \to 0} \frac{8}{x^{8}} \{1 - \cos \frac{x^{2}}{2} - \cos \frac{x^{2}}{4} + \cos \frac{x^{2}}{2} \cos \frac{x^{2}}{4}\}$ is equal to

(a) $\frac{1}{16}$

(b) $- \frac{1}{16}$

(c) $\frac{1}{32}$

(d) $- \frac{1}{32}$

Answer:

(c) $\frac{1}{32}$
$\lim_{x \to 0} \frac{8}{x^{8}} [1 - \cos \frac{x^{2}}{2} - \cos \frac{x^{2}}{4} + \cos \frac{x^{2}}{2} \cos \frac{x^{2}}{4}] = \lim_{x \to 0} \frac{8}{x^{8}} [(1 - \cos \frac{x^{2}}{4}) - \cos \frac{x^{2}}{2} (1 - \cos \frac{x^{2}}{4})] = \lim_{x \to 0} \frac{8}{x^{8}} [(1 - \cos \frac{x^{2}}{4}) (1 - \cos \frac{x^{2}}{2})] = \lim_{x \to 0} \frac{8}{x^{8}} [(2 \sin^{2} \frac{x^{2}}{8}) (2 \sin^{2} \frac{x^{2}}{4})] = \lim_{x \to 0} 4 \times 8 \frac{(\sin^{2} \frac{x^{2}}{8})}{(64 \times \frac{x^{4}}{64})} \frac{(\sin^{2} \frac{x^{2}}{4})}{16 (\frac{x^{4}}{16})} = \frac{32}{64 \times 16} = \frac{1}{32}$

Page No 29.79:

Question 30:

If α is a repeated root of ax² + bx + c = 0, then $\lim_{x \to α} \frac{\tan (a x^{2} + b x + c)}{{(x - α)}^{2}}$ is
(a) a
(b) b
(c) c
(d) 0

Answer:

(a)

$\lim_{x \to α} [\frac{\tan (a x^{2} + b x + c)}{{(x - α)}^{2}}] = \lim_{x \to α} [\frac{\tan \{a (x - α) (x - α)\}}{a {(x - α)}^{2}}] \times a = \lim_{x \to α} [\frac{\tan \{a {(x - α)}^{2}\}}{a {(x - α)}^{2}}] \times a = 1 \times a [\lim_{θ \to 0} (\frac{\tan θ}{θ}) = 1] = a$

Page No 29.80:

Question 31:

The value of $\lim_{x \to 0} \frac{\sqrt{a^{2} - a x + x^{2}} - \sqrt{a^{2} + a x + x^{2}}}{\sqrt{a + x} - \sqrt{a - x}}$ is

(a) a

(b) $\sqrt{a}$

(c) −a

(d) $- \sqrt{a}$

Answer:

(d) $- \sqrt{a}$

$\lim_{x \to 0} \frac{\sqrt{a^{2} - a x + x^{2}} - \sqrt{a^{2} + a x + x^{2}}}{\sqrt{a + x} - \sqrt{a - x}} Multiplying (\sqrt{a^{2} - a x + x^{2}} + \sqrt{a^{2} + a x + x^{2}}) and (\sqrt{a + x} + \sqrt{a - x}) in N^{r} and D^{r} : \lim_{x \to 0} \frac{(a^{2} + x^{2} - a x - a^{2} - a x - x^{2}) (\sqrt{a + x} + \sqrt{a - x})}{(a + x - a + x) (\sqrt{x^{2} - a x + x^{2}} + \sqrt{a^{2} + a x + x^{2}})} \lim_{x \to 0} \frac{- 2 a x (\sqrt{a + x} + \sqrt{a - x})}{2 x (\sqrt{a^{2} + a x + x^{2}} + \sqrt{a^{2} + a x + x^{2}})} = \frac{- a (2 \sqrt{a})}{2 a} = - \sqrt{a}$

Page No 29.80:

Question 32:

The value of $\lim_{x \to 0} \frac{1 - \cos x + 2 \sin x - \sin^{3} x - x^{2} + 3 x^{4}}{\tan^{3} x - 6 \sin^{2} x + x - 5 x^{3}}$ is

(a) 1

(b) 2

(c) −1

(d) −2

Answer:

(b) 2

$\lim_{x \to 0} \frac{(1 - \cos x) + 2 \sin x - \sin^{3} x - x^{2} + 3 x^{4}}{\tan^{3} x - 6 \sin^{2} x + x - 5 x^{3}} Divi ding N^{r} and D^{r} by x : \lim_{x \to 0} \frac{\frac{2 \sin^{2} \frac{x}{2}}{x} + \frac{2 \sin x}{x} - \frac{\sin^{3} x}{x} - x + 3 x^{3}}{\frac{\tan^{3} x}{x} - \frac{6 \sin^{2} x}{x} + 1 - 5 x^{2}} \lim_{x \to 0} \frac{\frac{2 x}{4} \frac{\sin^{2} \frac{x}{2}}{\frac{x^{2}}{4}} + \frac{2 \sin x}{x} - \frac{x^{2} \sin^{3} x}{x^{3}} - x + 3 x^{3}}{x^{2} (\frac{\tan 3 x}{x^{3}}) - \frac{6 x \sin^{2} x}{x^{2}} + 1 - 5 x^{2}} = \frac{0 + 2 - 0 - 0 + 0}{0 - 0 + 1 - 0} = 2$

Page No 29.80:

Question 33:

$\lim_{θ \to π / 2} \frac{1 - \sin θ}{(π / 2 - θ) \cos θ}$ is equal to

(a) 1

(b) −1

(c) $\frac{1}{2}$

(d) $- \frac{1}{2}$

Answer:

(c) $\frac{1}{2}$

$\lim_{θ \to \frac{π}{2}} \frac{1 - \sin θ}{(\frac{π}{2} - θ) \cos θ} = \lim_{h \to 0} \frac{1 - \cos h}{(\frac{π}{2} - (\frac{π}{2} - h)) \sin h} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{h \sin h} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{\frac{\frac{4 h^{2}}{4}}{\frac{\sin h}{h}}} = \frac{2}{4} = \frac{1}{2}$

Page No 29.80:

Question 34:

The value of $\lim_{x \to π / 2} (\sec x - \tan x)$ is

(a) 2

(b) −1

(c) 1

(d) 0

Answer:

(d) 0

$\lim_{x \to \frac{π}{2}} (\sec x - \tan x) = \lim_{h \to 0} (\sec (\frac{π}{2} - h) - \tan (\frac{π}{2} - h)) = \lim_{h \to 0} (cosec h - \cot h) = \lim_{h \to 0} \frac{1 - \cos h}{\sin h} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{\sin h} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{2 \sin \frac{h}{2} \cos \frac{h}{2}} = \lim_{h \to 0} \tan \frac{h}{2} = 0$

Page No 29.80:

Question 35:

The value of $\lim_{x \to \infty} \frac{n!}{(n + 1)! - n!}$ is

(a) 0

(b) 1

(c) −1

(d) none of these

Answer:

(a) 0
$\lim_{n \to \infty} \frac{n!}{(n + 1)! - n!} Divi ding N^{r} and D^{r} by n! : \lim_{n \to \infty} \frac{1}{\frac{(n + 1)!}{n!} - 1} = \lim_{n \to \infty} \frac{1}{\frac{(n + 1) n!}{n!} - 1} = \lim_{n \to \infty} \frac{1}{n + 1 - 1} = \lim_{n \to \infty} = \frac{1}{n} = 0$

Page No 29.80:

Question 36:

The value of $\lim_{n \to \infty} \frac{(n + 2)! + (n + 1)!}{(n + 2)! - (n + 1)!}$ is

(a) 0
(b) −1
(c) 1
(d) none of these

Answer:

(c) 1
$\lim_{n \to \infty} \frac{(n + 2)! + (n + 1)!}{(n + 2)! - (n + 1)!} Dividing N^{r} & D^{r} by (n + 1)! : \Rightarrow \lim_{n \to \infty} \frac{\frac{(n + 2) (n + 1)!}{(n + 1)!} + 1}{\frac{(n + 2) (n + 1)!}{(n + 1)!} - 1} = \lim_{n \to \infty} \frac{n + 2 + 1}{n + 2 - 1} = \lim_{n \to \infty} \frac{n + 3}{n + 1} = \lim_{n \to \infty} \frac{1 + \frac{3}{n}}{1 + \frac{1}{n}} = 1$

Page No 29.80:

Question 37:

The value of $\lim_{x \to \infty} \frac{{(x + 1)}^{10} + {(x + 2)}^{10} + . . . + {(x + 100)}^{10}}{x^{10} + 10^{10}}$ is

(a) 10

(b) 100

(c) 10¹⁰

(d) none of these

Answer:

(b) 100

$\lim_{x \to \infty} \frac{{(x + 1)}^{10} + {(x + 2)}^{10} + . . . . + {(x + 100)}^{10}}{x^{10} + 10^{10}} Dividing N^{r} and D^{r} by x^{10} : \Rightarrow \lim_{x \to \infty} \frac{{(1 + \frac{1}{x})}^{10} + {(1 + \frac{2}{x})}^{10} + . . . . + {(1 + \frac{100}{x})}^{10}}{1 + \frac{10^{10}}{x^{10}}} = 1 + 1 + 1 + . . . + 100 times = 100$

Page No 29.80:

Question 38:

The value of $\lim_{n \to \infty} \{\frac{1 + 2 + 3 + . . . + n}{n + 2} - \frac{n}{2}\}$ is

(a) 1/2

(b) 1

(c) −1

(d) −1/2

Answer:

(d) −1/2

$\lim_{n \to \infty} [\frac{1 + 2 + 3 + . . . . . n}{n + 2} - \frac{n}{2}] = \lim_{n \to \infty} [\frac{n (n + 1)}{2 (n + 2)} - \frac{n}{2}] = \lim_{n \to \infty} \frac{n}{2} [\frac{n + 1 - n - 2}{n + 2}] = \lim_{n \to \infty} \frac{n}{2} (\frac{- 1}{n + 2}) = \lim_{n \to \infty} \frac{- 1}{2 (1 + \frac{2}{n})} = \frac{- 1}{2 (1 + 0)} = - \frac{1}{2}$

Page No 29.80:

Question 39:

$\lim_{x \to 1} [x - 1]$ , where [.] is the greatest integer function, is equal to

(a) 1 (b) 2 (c) 0 (d) does not exist

Answer:

We have,

$\lim_{x \to 1^{-}} [x - 1] = \lim_{h \to 0} [1 - h - 1] = \lim_{h \to 0} [- h] = - 1$ (k − 1 < k − h < k ⇒ [k − h] = k − 1, k ∈ Z)

Also,

$\lim_{x \to 1^{+}} [x - 1] = \lim_{h \to 0} [1 + h - 1] = \lim_{h \to 0} [h] = 0$

$∴ \lim_{x \to 1^{-}} [x - 1] \neq \lim_{x \to 1^{+}} [x - 1]$

Thus, $\lim_{x \to 1} [x - 1]$ does not exist.

Hence, the correct answer is option (d).

Page No 29.80:

Question 40:

$\lim_{x \to \infty} \frac{|x|}{x}$ is equal to

(a) 1 (b) −1 (c) 0 (d) does not exist

Answer:

We know that,

$|x| = \{\begin{cases} x, & if x \geq 0 \\ - x, & if x < 0 \end{cases} ∴ \frac{|x|}{x} = \{\begin{cases} \frac{x}{x}, & if x \geq 0 \\ \frac{- x}{x}, & if x < 0 \end{cases} = \{\begin{cases} 1, & if x \geq 0 \\ - 1, & if x < 0 \end{cases}$

Now, for all x ≥ 0 (however, x may large be), $\frac{|x|}{x} = 1$ .

$∴ \lim_{x \to \infty} \frac{|x|}{x} = 1$

Hence, the correct answer is option (a).

Page No 29.80:

Question 41:

$\lim_{x \to 0} \frac{|\sin x|}{x}$ is

(a) 1 (b) −1 (c) 0 (d) does not exist

Answer:

We have,

$|\sin x| = \{\begin{cases} \sin x, & 0 \leq x \leq \frac{π}{2} \\ - \sin x, & - \frac{π}{2} \leq x < 0 \end{cases}$

Now,

$\lim_{x \to 0^{-}} \frac{|\sin x|}{x} = \lim_{x \to 0} \frac{- \sin x}{x} = - \lim_{x \to 0} \frac{\sin x}{x} = - 1$

$\lim_{x \to 0^{+}} \frac{|\sin x|}{x} = \lim_{x \to 0} \frac{\sin x}{x} = 1$

Clearly, $\lim_{x \to 0^{-}} \frac{|\sin x|}{x} \neq \lim_{x \to 0^{+}} \frac{|\sin x|}{x}$

∴ $\lim_{x \to 0} \frac{|\sin x|}{x}$ does not exist.

Hence, the correct answer is option (d).

Note: Here, a small interval close to 0 is only considered.

Page No 29.80:

Question 42:

If $f (x) = \{\begin{cases} \frac{\sin [x]}{[x]}, & [x] \neq 0 \\ 0, & [x] = 0 \end{cases}$ , where [.] denotes the greatest integer function, then $\lim_{x \to 0} f (x)$ is equal to

(a) 1 (b) 0 (c) −1 (d) does not exist

Answer:

We have,

$[x] = \{\begin{cases} 0, & 0 \leq x < 1 \\ - 1, & - 1 \leq x < 0 \end{cases}$

$∴ f (x) = \{\begin{cases} \frac{\sin (- 1)}{- 1}, & - 1 \leq x < 0 \\ 0, & 0 \leq x < 1 \end{cases} \Rightarrow f (x) = \{\begin{cases} \sin 1, & - 1 \leq x < 0 \\ 0, & 0 \leq x < 1 \end{cases}$

Now,

$\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0} \sin 1 = \sin 1 \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0} 0 = 0$

Clearly, $\lim_{x \to 0^{-}} f (x) \neq \lim_{x \to 0^{+}} f (x)$

Thus, $\lim_{x \to 0} f (x)$ does not exist.

Hence, the correct answer is option (d).

Page No 29.80:

Question 43:

$\lim_{x \to π} \frac{\sin x}{x - π} is$

(a) 1
(b) 2
(c) –1
(d) –2

Answer:

$\begin{matrix} \lim \\ x \to π \end{matrix} \frac{\sin x}{x - π} put x - π = h i . e . x = π + h where h \to 0 . = \begin{matrix} \lim \\ h \to 0 \end{matrix} \frac{\sin (π + h)}{(π + h - π)} = \begin{matrix} \lim \\ h \to 0 \end{matrix} \frac{(- \sin h)}{h} = - (\begin{matrix} \lim \\ h \to 0 \end{matrix} \frac{\sin h}{h}) [∵ \frac{sinx}{x} x \to 0 1] = - 1 . Hence, the correct answer is option C .$

Page No 29.81:

Question 44:

$\lim_{x \to 0} \frac{x^{2} \cos x}{1 - \cos x} is$

(a) 2

(b) $\frac{3}{2}$

(c) $- \frac{3}{2}$

(d) 1

Answer:

$\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{x^{2} \cos x}{1 - \cos x} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} (\frac{x^{2}}{2 \sin^{2} \frac{x}{2}}) \cos x (∵ 1 - \cos 2 θ = 2 \sin^{2} θ) i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{1}{2} (\frac{2^{2} {(\frac{x}{2})}^{2}}{\sin^{2} \frac{x}{2}}) \cos x i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} 2 {(\frac{\frac{x}{2}}{\sin \frac{x}{2}})}^{2} \cos x i . e . \begin{matrix} 2 \lim \\ x \to 0 \end{matrix} {(\frac{\frac{x}{2}}{\sin \frac{x}{2}})}^{2} (\begin{matrix} \lim \\ x \to 0 \end{matrix} \cos x) = 2 \times 1 \times 1 = 2 Hence, the correct answer is option A .$

Page No 29.81:

Question 45:

$\lim_{x \to 0} \frac{{(1 - x)}^{n} - 1}{x} is$

(a) n
(b) 1
(c) –n
(d) 0

Answer:

$\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{{(1 + x)}^{n} - 1}{x} since {(1 + x)}^{n} = 1 + n x + \frac{n (n - 1)}{2} x^{2} + . . . . . . . . . . . . . . . . . . + x^{n} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{[1 + n x + \frac{n (n - 1) x^{2}}{2} + . . . . . . . . . . . . . . . . . . + x^{n}] - 1}{x} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{x [n + \frac{n (n - 1)}{2} x + . . . . . . . . . . . . . . . . . . + x^{n - 1}]}{x} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} [n + \frac{n (n - 1) x}{2} + . . . . . . . . . . . . . . . . . . + x^{n - 1}] i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} n + \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{n (n - 1) x}{2} + . . . . . . . . . . . . . . . . . . + \begin{matrix} \lim \\ x \to 0 \end{matrix} x^{n} = n$

Hence, the correct answer is option A.

Page No 29.81:

Question 46:

$\lim_{x \to 1} \frac{x^{m} - 1}{x^{n} - 1} is$

(a) 1

(b) $\frac{m}{n}$

(c) $- \frac{m}{n}$

(d) $\frac{m^{2}}{n^{2}}$

Answer:

$\begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{x^{m} - 1}{x^{n} - 1} since \frac{x^{n} - a^{n}}{x - a} \overset{x + a}{\to} n a^{n - 1} i . e . \begin{matrix} \lim \\ x \to a \end{matrix} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1} i . e . \begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{x^{m} - 1}{x - 1} \times \frac{x - 1}{x^{n} - 1} i . e . (\begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{x^{m} - 1}{x - 1}) (\begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{x - 1}{x^{n} - 1}) = m 1 \times \frac{1}{n} = 1 \times \frac{m}{n} = \frac{m}{n}$

Hence, the correct answer is option B.

Page No 29.81:

Question 47:

$\lim_{θ \to 0} \frac{1 - \cos 4 θ}{1 - \cos 6 θ} is$

(a) $\frac{4}{9}$

(b) $\frac{1}{2}$

(c) $- \frac{1}{2}$

(d) –1

Answer:

$\begin{matrix} \begin{matrix} \lim \\ θ \to 0 \end{matrix} \end{matrix} \frac{1 - \cos 4 θ}{1 - \cos 6 θ} (since 1 - \cos 2 x = 2 \sin^{2} x) i . e . \begin{matrix} \lim \\ θ \to 0 \end{matrix} \frac{2 \sin^{2} 2 θ}{2 \sin^{2} 3 θ} i . e . \begin{matrix} \lim \\ θ \to 0 \end{matrix} \frac{\sin^{2} 2 θ}{\sin^{2} 3 θ} = \begin{matrix} \lim \\ θ \to 0 \end{matrix} \frac{\sin^{2} 2 θ}{{(2 θ)}^{2}} \times \frac{{(3 θ)}^{2}}{\sin^{2} 3 θ} \times \frac{{(2 θ)}^{2}}{{(3 θ)}^{2}} = \begin{matrix} \lim \\ θ \to 0 \end{matrix} \frac{\sin^{2} 2 θ}{{(2 θ)}^{2}} \times \frac{{(3 θ)}^{2}}{\sin^{2} 3 θ} \times \frac{4}{9} = \frac{4}{9} (\begin{matrix} \lim \\ θ \to 0 \end{matrix} \frac{\sin^{2} 2 θ}{{(2 θ)}^{2}}) (\begin{matrix} \lim \\ θ \to 0 \end{matrix} \frac{{(3 θ)}^{2}}{\sin^{2} 3 θ}) i . e . \begin{matrix} \lim \\ θ \to 0 \end{matrix} \frac{1 - \cos 4 θ}{1 - \cos 6 θ} = \frac{4}{9} \times 1 \times 1 = \frac{4}{9}$

Hence, the correct answer is option A.

Page No 29.81:

Question 48:

$\lim_{x \to 0} \frac{\cos e c x - \cot x}{x} is$

(a) $- \frac{1}{2}$

(b) 1

(c) $\frac{1}{2}$

(d) 1

Answer:

$\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{cosec x - \cot x}{x} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\frac{1}{\sin x} - \frac{\cos x}{\sin x}}{x} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{1 - \cos x}{x \sin x} [since 1 - \cos 2 θ = 2 \sin^{2} θ and \sin 2 θ = 2 \sin θ \cos θ] i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{2 \sin^{2} \frac{x}{2}}{x 2 \cos \frac{x}{2} \sin \frac{x}{2}} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\sin \frac{x}{2}}{x} \times \frac{1}{\cos \frac{x}{2}} \begin{matrix} \lim \\ x \to 0 \end{matrix} (\frac{\sin \frac{x}{2}}{2 \frac{x}{2}}) (\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{1}{\cos \frac{x}{2}}) = \frac{1}{2} (\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\sin \frac{x}{2}}{\frac{x}{2}}) (\frac{1}{1}) = \frac{1}{2} (∵ \begin{matrix} \lim \\ θ \to 0 \end{matrix} \frac{\sin θ}{θ} = 1)$

Hence,the correct answer is option C.

Page No 29.81:

Question 49:

$\lim_{x \to π / 4} \frac{\sec^{2} x - 2}{\tan x - 1} is$

(a) 3
(b) 1
(c) 2
(d) $\sqrt{2}$

Answer:

$\begin{matrix} \lim \\ x \to \frac{π}{4} \end{matrix} \frac{\sec^{2} x - 1 - 1}{\tan x - 1} (since \sec^{2} x - 1 = \tan^{2} x) i . e . \begin{matrix} \lim \\ x \to \frac{π}{4} \end{matrix} \frac{\tan^{2} x - 1}{\tan x - 1} i . e . \begin{matrix} \lim \\ x \to \frac{π}{4} \end{matrix} \frac{(\tan x - 1) (\tan x + 1)}{\tan x - 1} i . e . \begin{matrix} \lim \\ x \to \frac{π}{4} \end{matrix} \tan x + 1 = 1 + 1 = 2 i . e . \begin{matrix} \lim \\ x \to \frac{π}{4} \end{matrix} \frac{\sec^{2} x - 2}{\tan x - 1} = 2$

Hence, the correct answer is option C.

Page No 29.81:

Question 50:

$\lim_{x \to 0} \frac{\sin x}{\sqrt{1 + x} - \sqrt{1 - x}} is$

(a) 2
(b) 0
(c) 1
(d) –1

Answer:

$\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\sin x}{\sqrt{1 + x} - \sqrt{1 - x}} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\sin x}{\sqrt{1 + x} - \sqrt{1 - x}} \times \frac{(\sqrt{1 + x} + \sqrt{1 - x})}{(\sqrt{1 + x} + \sqrt{1 - x})} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\sin x}{{(\sqrt{1 + x})}^{2} - {(\sqrt{1 - x})}^{2}} \times \sqrt{1 + x} + \sqrt{1 - x} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\sin x}{1 + x - 1 + x} \times \sqrt{1 + x} + \sqrt{1 - x} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\sin x}{2 x} \times \sqrt{1 + x} + \sqrt{1 - x} i . e . \frac{1}{2} (\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\sin x}{x}) (\begin{matrix} \lim \\ x \to 0 \end{matrix} \sqrt{1 + x} + \sqrt{1 - x}) i . e . \frac{1}{2} \times 1 \times (\sqrt{1} + \sqrt{1}) i . e . \frac{1}{2} \times 2 = 1$

Hence, the correct answer is option C.

Page No 29.81:

Question 51:

$\lim_{x \to 1} \frac{(\sqrt{x} - 1) (2 x - 3)}{2 x^{2} + x - 3} is$

$(a) \frac{1}{10} (b) - \frac{1}{10}$

(c) 1

(d) none of these

Answer:

$i . e . \begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{(\sqrt{x} - 1) (2 x - 3)}{2 x^{2} + x - 3} i . e . \begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{(\sqrt{x} - 1) (2 x - 3)}{2 x^{2} + 3 x - 2 x - 3} i . e . \begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{(\sqrt{x} - 1) (2 x - 3)}{(2 x^{2} - 2 x) + (3 x - 3)} i . e . \begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{(\sqrt{x} - 1) (2 x - 3)}{2 x (x - 1) + 3 (x - 1)} i . e . \begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{(\sqrt{x} - 1) (2 x - 3)}{(x - 1) (2 x + 3)} i . e . \begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{(\sqrt{x} - 1) (2 x - 3)}{(\sqrt{x} - 1) (\sqrt{x} + 1) (2 x + 3)} i . e . \begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{2 x - 3}{(\sqrt{x} + 1) (2 x + 3)} i . e . (\begin{matrix} \lim \\ x \to 1 \end{matrix} 2 x - 3) (\begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{1}{\sqrt{x} + 1}) (\begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{1}{2 x + 3}) i . e . (- 1) \times \frac{1}{2} \times \frac{1}{5} = \frac{- 1}{10}$

Hence, the correct answer is option B.

Page No 29.82:

Question 1:

If f(x) $\frac{\tan x}{x - π}, then \lim_{x \to π}$ f(x) = _____________________________.

Answer:

$Given f (x) = \frac{\tan x}{x - π} \begin{matrix} \begin{matrix} \lim \\ x \to π \end{matrix} \end{matrix} f (x) = \begin{matrix} \lim \\ x \to π \end{matrix} \frac{\tan x}{x - π} put x - π = h i . e . x = h + π where h \to 0 i . e . \begin{matrix} \lim \\ h \to 0 \end{matrix} \frac{\tan (π + h)}{h} i . e . \begin{matrix} \lim \\ h \to 0 \end{matrix} \frac{\tan h}{h} (∵ \tan (π + θ) = tanθ) = 1 (∵ \begin{matrix} \lim \\ θ \to 0 \end{matrix} \frac{tanθ}{θ} = 1) Hence, \begin{matrix} \lim \\ x \to π \end{matrix} f (x) = 1 .$

Page No 29.82:

Question 2:

$\lim_{x \to 3^{-}} \frac{x}{[x]} =__________________.$

Answer:

$\begin{matrix} \lim \\ x \to 3^{-} \end{matrix} \frac{x}{[x]} i . e . \begin{matrix} \lim \\ h \to 0 \end{matrix} \frac{x - h}{[3 - h]} for \begin{matrix} \lim \\ x \to 3 \end{matrix} x - h i . e . \begin{matrix} \lim \\ h \to 0 \end{matrix} 3 - h i . e . \begin{matrix} \lim \\ h \to 0 \end{matrix} \frac{3 - h}{[3 - h]} = \begin{matrix} \lim \\ h \to 0 \end{matrix} \frac{3 - h}{3} = \begin{matrix} \lim \\ h \to 0 \end{matrix} 1 - \frac{h}{3} = 1 Hence, \begin{matrix} \lim \\ x \to 3^{-} \end{matrix} \frac{x}{[3]} = 1 .$

Page No 29.82:

Question 3:

$If \lim_{x \to 0} (\sin m x \cot \frac{x}{\sqrt{3}}) = 2, then m =______________________________.$

Answer:

$\begin{matrix} \lim \\ x \to 0 \end{matrix} (\sin m x \cot \frac{x}{\sqrt{3}}) = 2 i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} (\frac{\sin m x}{m x} \times m x \times \frac{1}{\tan \frac{x}{\sqrt{3}}}) = 2 i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} (\frac{\sin m x}{m x} \times m x \times \frac{\frac{x}{\sqrt{3}}}{\tan \frac{x}{\sqrt{3}}} \times \frac{1}{\frac{x}{\sqrt{3}}}) = 2 i . e . (\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\sin m x}{m x}) (\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{m x}{\frac{x}{\sqrt{3}}}) (\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\frac{x}{\sqrt{3}}}{\tan \frac{x}{\sqrt{3}}}) = 2 i . e . 1 \times m \sqrt{3} \times 1 = 2 i . e . m = \frac{2}{\sqrt{3}}$

Page No 29.82:

Question 4:

$\lim_{x \to 0} \frac{|\sin x|}{x}________________________________.$

Answer:

$\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{|\sin x|}{x} L . H . L \begin{matrix} \lim \\ x \to 0^{-} \end{matrix} \frac{|\sin x|}{x} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{- \sin x}{x} i . e . \begin{matrix} - \lim \\ x \to 0 \end{matrix} \frac{\sin x}{x} = - 1 i . e . left hand limit is - 1 R . H . L \begin{matrix} \lim \\ x \to 0^{+} \end{matrix} \frac{|\sin x|}{x} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\sin x}{x} i . e . R . H . L = 1 Since L . H . L \neq R . H . S Hence, \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{|\sin x|}{x} does not exist$

Page No 29.82:

Question 5:

$\lim_{x \to \frac{π}{4}} \frac{{cosec}^{2} x - 2}{\cot x - 1}$ is equal to ______________________.

Answer:

$\begin{matrix} \lim \\ x \to \frac{π}{4} \end{matrix} \frac{{cosec}^{2} x - 2}{\cot x - 1} i . e . \begin{matrix} \lim \\ x \to \frac{π}{4} \end{matrix} \frac{{cosec}^{2} x - 1 - 1}{\cot x - 1} (since {cosec}^{2} x - 1 = \cos^{2} x) i . e . \begin{matrix} \lim \\ x \to \frac{π}{4} \end{matrix} \frac{\cot^{2} x - 1}{\cot x - 1} i . e . \begin{matrix} \lim \\ x \to \frac{π}{4} \end{matrix} \frac{(\cot x - 1) (\cot x + 1)}{\cot x - 1} i . e . \begin{matrix} \lim \\ x \to \frac{π}{4} \end{matrix} \cot x + 1 = 1 + 1 = 2 Hence, \begin{matrix} \lim \\ x \to \frac{π}{4} \end{matrix} \frac{{cosec}^{2} x - 2}{\cot x - 1} = 2$

Page No 29.82:

Question 6:

$\lim_{x \to 1} \frac{x^{m} - 1}{x^{n} - 1}$ is equal to ______________________.

Answer:

$\begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{x^{m} - 1}{x^{n} - 1} i . e . \begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{x^{m} - 1}{x - 1} \times \frac{x - 1}{x^{n} - 1} i . e . (\begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{x^{m} - 1}{x - 1}) (\begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{x - 1}{x^{n} - 1}) i . e . m \times \frac{1}{n} (∵ \begin{matrix} \lim \\ x \to a \end{matrix} \frac{x^{n} - a}{x - a} = n a^{n - 1}) i . e . \frac{m}{n} i . e . \begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{x^{m} - 1}{x^{n} - 1} = \frac{m}{n}$

Page No 29.82:

Question 7:

$\lim_{n \to \infty} \frac{1 + 2 + 3 + . . . + n}{n^{2} + 100}_________________________.$

Answer:

$\begin{matrix} \lim \\ n \to \infty \end{matrix} \frac{1 + 2 + . . . . . . . . . . + n}{n^{2} + 100} i . e . \begin{matrix} \lim \\ n \to \infty \end{matrix} \frac{\frac{n (n + 1)}{2}}{n^{2} + 100} (∵ Sum of first n natural number is \frac{n (n + 1)}{2}) i . e . \begin{matrix} \lim \\ n \to \infty \end{matrix} \frac{1}{2} [\frac{n^{2} + n}{n^{2} + 100}] i . e . \begin{matrix} \lim \\ n \to \infty \end{matrix} \frac{1}{2} [\frac{1 + \frac{1}{n}}{1 + \frac{100}{n^{2}}}] (By dividing by n^{2} both numerator of denominator) (Since \begin{matrix} \lim \\ n \to \infty \end{matrix} \frac{1}{n} = 0)$
$\begin{array}{rcl} i . e . \begin{matrix} \lim \\ n \to \infty \end{matrix} \frac{1}{2} (\frac{n (n + 1)}{n^{2} + 100}) & = & \frac{1}{2} \frac{(1 + \begin{matrix} \lim \\ n \to \infty \end{matrix} \frac{1}{n})}{1 + \begin{matrix} \lim \\ n \to \infty \end{matrix} \frac{100}{n^{2}}} \\ = & \frac{1}{2} \frac{(1 + 0)}{1 + 0} \end{array}$
$i . e . \begin{matrix} \lim \\ n \to \infty \end{matrix} \frac{1 + 2 + . . . . . + n}{n^{2} + 100} = \frac{1}{2}$

Page No 29.82:

Question 8:

$\lim_{x \to 0} \frac{e^{2 \tan x} - 1}{x}$ is equal to _____________________.

Answer:

$\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{e^{2 \tan x} - 1}{x} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{e^{2 \tan x} - 1}{2 \tan x} \times \frac{2 \tan x}{x} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} (\frac{e^{2 \tan x} - 1}{2 \tan x}) \times 2 (\frac{\tan x}{x}) i . e . (\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{e^{2 \tan x} - 1}{2 \tan x}) \times 2 (\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\tan x}{x}) i . e . 1 \times 2 \times 1 = 2 i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{e^{2 \tan x} - 1}{x} = 2$

Page No 29.82:

Question 9:

$\lim_{x \to 1} \frac{\log_{e} x}{x - 1}$ is equal to __________________.

Answer:

$\begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{\log_{e} x}{x - 1} i . e . \begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{\log_{e} (x + 1 - 1)}{x - 1} i . e . \begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{\log [1 + (x - 1)]}{x - 1} = 1 (∵ \log \frac{1 + θ}{θ} \to 1 as θ + θ) i . e . \begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{\log_{e} x}{x - 1} = 1 .$

Page No 29.82:

Question 10:

Let f (x) = $\{\begin{cases} x^{2} - 1, 0 < x < 2 \\ 2 x + 3, 2 \leq x < 3 \end{cases}$ , the quadratic equation whose roots are $\lim_{x \to 2^{-}}$ f (x) and $\lim_{x \to 2^{+}}$ f (x) is __________________.

Answer:

$f (x) = \{\begin{array}{l} x^{2} - 1; \\ 2 x + 3; \end{array} \begin{array}{r} 0 < x < 2 \\ 2 \leq x < 3 \end{array}\} Given Then,$

$\begin{array}{rcl} \begin{matrix} \lim \\ x \to 2^{-} \end{matrix} f (x) & = & \begin{matrix} \lim \\ x \to 2 \end{matrix} x^{2} - 1 \\ = & {(2)}^{2} - 1 = 4 - 1 = 3 \end{array}$

$and \begin{matrix} \lim \\ x \to 2^{+} \end{matrix} f (x) = \begin{matrix} \lim \\ x \to 2 \end{matrix} 2 x + 3 = 2 (2) + 3 = 7 Hence, equation where roots are \begin{matrix} \lim \\ x \to 2^{-} \end{matrix} f (x) and \begin{matrix} \lim \\ x \to 2^{+} \end{matrix} f (x) is (x - 3) (x - 7) = x^{2} - 10 x + 21 = 0$

Page No 29.82:

Question 11:

$\lim_{x \to 1} \frac{\tan (x^{2} - 1)}{x - 1}$ is equal to _____________________.

Answer:

$\begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{\tan (x^{2} - 1)}{x - 1} = \begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{\tan (x^{2} - 1)}{x - 1} \times \frac{x + 1}{x + 1} = \begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{\tan (x^{2} - 1)}{(x^{2} - 1)} \times x + 1 i . e . (\begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{\tan (x^{2} - 1)}{(x^{2} - 1)}) (\begin{matrix} \lim \\ x \to 1 \end{matrix} x + 1) (since \begin{matrix} \lim \\ θ \to 0 \end{matrix} \frac{\tan θ}{θ} = 1) i . e . 1 \times (1 + 1) = 2 i . e . \begin{matrix} \lim \\ x \to 1 \end{matrix} \frac{\tan (x^{2} - 1)}{x - 1} = 2$

Page No 29.82:

Question 12:

$\lim_{x \to 0} \frac{\tan x^{0}}{x}$ is equal to _______________________.

Answer:

$\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\tan x °}{x} since x ° = \frac{π}{180} x i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\tan \frac{π}{180} x}{x \times \frac{π}{180}} \times \frac{π}{180} i . e . (\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\tan \frac{π}{180} x}{\frac{π}{180} x}) (\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{π}{180}) i . e . 1 \times \frac{π}{180} i . e . \frac{π}{180} Hence, \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\tan x °}{x} = \frac{π}{180}$

Page No 29.82:

Question 13:

$\lim_{x \to 0} \frac{2 x + 3 \tan x}{3 x - 2 \sin x}$ is equal to_________________________.

Answer:

$\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{2 x + 3 \tan x}{3 x - 2 \sin x} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{2 + 3 \frac{\tan x}{x}}{3 - 2 \frac{\sin x}{x}} (dividing number and denominator by x) i . e . \frac{2 + 3 \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\tan x}{x}}{3 - 2 \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\sin x}{x}} = \frac{2 + 1 \times 3}{3 - 2 \times 1} = \frac{5}{1} = 5$

Page No 29.82:

Question 14:

$\lim_{x \to 0} \frac{5 x \cos x - 2 \sin x}{3 x + \tan x}$ ________________________.

Answer:

$\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\sqrt{x} \cos x - 2 \sin x}{3 x + \tan x} By dividing number and denominator by x, i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{5 \cos x - 2 \frac{\sin x}{x}}{3 + \frac{\tan x}{x}} i . e . \frac{5 (\begin{matrix} \lim \\ x \to 0 \end{matrix} \cos x) - 2 (\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\sin x}{x})}{3 + (\begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{\tan x}{x})} i . e . \frac{5 \times 1 - 2 \times 1}{3 + 1} = \frac{5 - 2}{4} = \frac{3}{4} i . e . \begin{matrix} \lim \\ x \to 0 \end{matrix} \frac{5 x \cos x - 2 \sin x}{3 x + \tan x} = \frac{3}{4}$

Page No 29.82:

Question 15:

$If f (x) = \{\begin{cases} x, x > 1 \\ x^{2}, x < 1 \end{cases}, then \lim_{x \to 2} f (x) =____________________________________.$

Answer:

$f (x) \{\begin{cases} x; x > 1 \\ x^{2}; x < 1 \end{cases} Hence, \begin{matrix} \lim \\ x \to 1 \end{matrix} f (x) for x < 1 i . e . L . H . L is given by \begin{matrix} \lim \\ x \to 1^{-} \end{matrix} f (x) = \begin{matrix} \lim \\ x \to 1 \end{matrix} x^{2} = {(1)}^{2} = 1 i . e . L . H . L \begin{matrix} \lim \\ x \to 1^{-} \end{matrix} f (x) = 1 and R . H . L \begin{matrix} \lim \\ x \to 1^{+} \end{matrix} f (x) = \begin{matrix} \lim \\ x \to 1 \end{matrix} x = 1 Hence L . H . L = R . H . L i . e . \begin{matrix} \lim \\ x \to 1 \end{matrix} f (x) = 1$

Page No 29.82:

Question 16:

$\lim_{x \to \infty} (\sqrt{x^{2} + 1} - x)$ is equal to _____________________________.

Answer:

$\begin{matrix} \lim \\ x \to \infty \end{matrix} (\sqrt{x^{2} + 1} - x) i . e . \begin{matrix} \lim \\ x \to \infty \end{matrix} \sqrt{x^{2} + 1} - x \times (\frac{\sqrt{x^{2} + 1} + x}{\sqrt{x^{2} + 1} + x}) i . e . \begin{matrix} \lim \\ x \to \infty \end{matrix} \frac{{(\sqrt{x^{2} + 1})}^{2} - x^{2}}{\sqrt{x^{2} + 1} + x} i . e . \begin{matrix} \lim \\ x \to \infty \end{matrix} \frac{x^{2} + 1 - x^{2}}{\sqrt{x^{2} + 1} + x} = \begin{matrix} \lim \\ x \to \infty \end{matrix} \frac{x^{2} + 1 - x^{2}}{\sqrt{x^{2} + 1} + x} i . e . \begin{matrix} \lim \\ x \to \infty \end{matrix} \frac{1}{\sqrt{x^{2} + 1} + x} = \frac{1}{\infty} = 0 i . e . \begin{matrix} \lim \\ x \to \infty \end{matrix} (\sqrt{x^{2} + 1} - x) = 0$

Page No 29.82:

Question 17:

The value of $\lim_{x \to 2} \frac{3^{x / 2} - 3}{3^{x} - 9} is___________________.$

Answer:

$\begin{matrix} \lim \\ x \to 2 \end{matrix} \frac{3^{\frac{x}{2}} - 3}{3^{x} - 9} i . e . \begin{matrix} \lim \\ x \to 2 \end{matrix} \frac{3^{\frac{x}{2}} - 3}{{(3^{\frac{x}{2}})}^{2} - {(3)}^{2}} i . e . \begin{matrix} \lim \\ x \to 2 \end{matrix} \frac{3^{\frac{x}{2}} - 3}{(3^{\frac{x}{2}} - 3) (3^{\frac{x}{2}} + 3)} i . e . \begin{matrix} \lim \\ x \to 2 \end{matrix} \frac{1}{3^{\frac{x}{2}} + 3} i . e . \begin{matrix} \lim \\ x \to 2 \end{matrix} \frac{1}{3^{\frac{x}{2}} + 3} i . e . \frac{1}{3^{\frac{2}{2}} + 3} i . e . \frac{1}{3 + 3} = \frac{1}{6} Hence, \begin{matrix} \lim \\ x \to 2 \end{matrix} \frac{3^{\frac{x}{2}} - 3}{3^{x} - 9} = \frac{1}{6}$

Page No 29.83:

Question 1:

Write the value of $\lim_{x \to 0} \frac{\sqrt{1 - \cos 2 x}}{x} .$

Answer:

$\lim_{x \to 0} (\frac{\sqrt{1 - \cos 2 x}}{x}) = \lim_{x \to 0} (\frac{\sqrt{2 \sin^{2} x}}{x}) = \sqrt{2} \lim_{x \to 0} (\frac{\sqrt{\sin^{2} x}}{x}) = \sqrt{2} \lim_{x \to 0} (\frac{|\sin x|}{x}) LHL : = \sqrt{2} \lim_{x \to 0^{-}} (\frac{|\sin x|}{x}) Let x = 0 - h, where h \to 0 . = \sqrt{2} \lim_{h \to 0} (\frac{|\sin (- h)|}{- h}) = \sqrt{2} \lim_{h \to 0} (\frac{\sin h}{- h}) = - \sqrt{2} RHL : = \sqrt{2} \lim_{x \to 0^{+}} (\frac{|\sin x|}{x}) Let x = 0 + h, where h \to 0 . = \sqrt{2} \lim_{h \to 0} \frac{|\sin h|}{h} = \sqrt{2} LHL \neq RHL Thus, \lim_{x \to 0} (\frac{\sqrt{1 - \cos 2 x}}{x}) does not exist .$

Page No 29.83:

Question 2:

Write the value of $\lim_{x \to 0^{-}} [x] .$

Answer:

$\lim_{x \to 0^{-}} [x] If x = 0 - h, then h \to 0 . \lim_{h \to 0} [0 - h] = - 1$

Page No 29.83:

Question 3:

Write the value of $\lim_{x \to 0^{+}} [x] .$

Answer:

$\lim_{x \to 0^{+}} [x] Let x = 0 + h, where h \to 0 . \lim_{h \to 0} [0 + h] = 0$
Disclaimer: The solution given in the book is incorrect. However, the solution given here has been created according to the question given in the book.

Page No 29.83:

Question 4:

Write the value of $\lim_{x \to 1^{-}} x - [x] .$

Answer:

$\lim_{x \to 1^{-}} (x - [x]) x = 1 - h ∴ h \to 0 \lim_{h \to 0} ((1 - h) - [1 - h]) = 1 - 0 = 1$

Page No 29.83:

Question 5:

Write the value of $\lim_{x \to 0^{-}} \frac{\sin [x]}{[x]} .$

Answer:

$\lim_{x \to 0^{-}} (\frac{\sin [x]}{[x]}) x = 0 - h ∴ h \to 0 = \lim_{h \to 0} (\frac{\sin [0 - h]}{[0 - h]}) = \frac{\sin (- 1)}{- 1} = \frac{- \sin 1}{- 1} = \sin 1$

Page No 29.83:

Question 6:

Write the value of $\lim_{x \to π} \frac{\sin x}{x - π} .$

Answer:

$\lim_{x \to π} (\frac{\sin x}{x - π}) LHL : \lim_{x \to π^{-}} (\frac{\sin x}{x - π}) If x = π - h, then h \to 0 . = \lim_{h \to 0} (\frac{\sin (π - h)}{π - h - π}) = \lim_{h \to 0} (\frac{\sin h}{- h}) = - 1 RHL : \lim_{x \to π^{+}} (\frac{\sin x}{x - π}) If x = π + h, then h \to 0 . = \lim_{h \to 0} (\frac{\sin (π + h)}{π + h - π}) = \lim_{h \to 0} (\frac{- \sin h}{h}) = - 1 ∴ \lim_{x \to π} (\frac{\sin x}{x - π}) = - 1$

Page No 29.83:

Question 7:

Write the value of $\lim_{x \to \infty} \frac{\sin x}{x} .$

Answer:

$\lim_{x \to \infty} (\frac{\sin x}{x}) Let x = \frac{1}{y} If x \to \infty, then y \to 0 . = \lim_{y \to 0} y \cdot \sin (\frac{1}{y}) LHL : Let y = 0 - h If y \to 0, then h \to 0 . = \lim_{h \to 0} ((0 - h) \times \sin (\frac{1}{0 - h})) 0 \times The oscillating number between -1 and 1 = 0 RHL : \lim_{y \to 0^{+}} (y \cdot \sin (\frac{1}{y})) Let y = 0 + h If y \to 0, then h \to 0 . = \lim_{h \to 0} h \times \sin (\frac{1}{h}) = 0 \times The oscillating number between –1 and 1 = 0$

Page No 29.83:

Question 8:

Write the value of $\lim_{x \to 2} \frac{|x - 2|}{x - 2} .$

Answer:

$\lim_{x \to 2} \frac{|x - 2|}{x - 2} LHL : \lim_{x \to 2^{-}} (\frac{|x - 2|}{x - 2}) Let x = 2 - h If x \to 2, then h \to 0 . = \lim_{h \to 0} \frac{|2 - h - 2|}{2 - h - 2} = - 1 RHL : \lim_{x \to 2^{+}} (\frac{|x - 2|}{x - 2}) Let x = 2 + h If x \to 2, then h \to 0 . \lim_{h \to 0} \frac{|2 + h - 2|}{2 + h - 2} = 1 LHL \neq RHL Thus, \lim_{x \to 2} (\frac{|x - 2|}{x - 2}) does not exist .$

Page No 29.83:

Question 9:

Write the value of $\lim_{x \to 0} \frac{\sin x^{\circ}}{x} .$

Answer:

$\lim_{x \to 0} (\frac{\sin x °}{x}) = \lim_{x \to 0} (\frac{\sin (\frac{π x}{180})}{\frac{π}{180} x}) \times \frac{π}{180} = 1 \times \frac{π}{180} = \frac{π}{180}$

Page No 29.83:

Question 10:

Write the value of $\lim_{x \to 0^{-}} \frac{\sin x}{\sqrt{x}} .$

Answer:

$\lim_{x \to 0^{-}} (\frac{\sin x}{\sqrt{x}}) Let x = 0 - h If x \to 0, then h \to 0 . \lim_{h \to 0} (\frac{\sin (- h)}{\sqrt{- h}}) The value does not exist.$

Page No 29.83:

Question 11:

Write the value of $\lim_{x \to 0} \frac{\sin x}{\sqrt{1 + x} - 1} .$

Answer:

$\lim_{x \to 0} [\frac{\sin x}{\sqrt{1 + x} - 1}] Rationalising the denominator, we get : = \lim_{x \to 0} [\frac{\sin x}{(\sqrt{1 + x} - 1)} \times (\frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1})] = \lim_{x \to 0} [\frac{\sin x \times (\sqrt{1 + x} + 1)}{1 + x - 1}] = \lim_{x \to 0} [(\frac{\sin x}{x}) (\sqrt{1 + x} + 1)] = 1 \times (\sqrt{1} + 1) = 2$

Page No 29.83:

Question 12:

Write the value of $\lim_{x \to - \infty} (3 x + \sqrt{9 x^{2} - x}) .$

Answer:

$\lim_{x \to - \infty} (3 x + \sqrt{9 x^{2} - x}) Let m = - x If x \to - \infty, then m \to \infty . = \lim_{m \to \infty} (- 3 m + \sqrt{9 m^{2} + m}) Rationalising the numerator, we get : = \lim_{m \to \infty} \frac{(- 3 m + \sqrt{9 m^{2} + m}) (- 3 m - \sqrt{9 m^{2} + m})}{- 3 m - \sqrt{9 m^{2} + m}} = \lim_{m \to \infty} (\frac{- [- 9 m^{2} + (9 m^{2} + m)]}{- 3 m - \sqrt{9 m^{2} + m}}) = \lim_{m \to \infty} (\frac{- m}{- 3 m - \sqrt{9 m^{2} + m}}) Dividing the numerator and the denominator by m, and applying limit, we get: = \frac{- 1}{- 3 - \sqrt{9 + \frac{1}{\infty}}} = \frac{1}{6}$

Page No 29.83:

Question 13:

Write the value of $\lim_{n \to \infty} \frac{n! + (n + 1)!}{(n + 1)! + (n + 2)!} .$

Answer:

$\lim_{n \to \infty} [\frac{n! + (n + 1)!}{(n + 1)! + (n + 2)!}] = \lim_{n \to \infty} [\frac{n! + (n + 1) n!}{(n + 1)! + (n + 2) (n + 1)!}] = \lim_{n \to \infty} [\frac{n! (1 + n + 1)}{(n + 1)! (1 + n + 2)}] = \lim_{n \to \infty} [\frac{n! (n + 2)}{(n + 1) n! (n + 3)}] = \lim_{n \to \infty} [\frac{n + 2}{(n + 1) (n + 3)}] D egree of the denominator is greater than that of the numerator. \Rightarrow 0$

Page No 29.83:

Question 14:

Write the value of $\lim_{x \to π / 2} \frac{2 x - π}{\cos x} .$

Answer:

$\lim_{x \to \frac{π}{2}} [\frac{2 x - π}{\cos x}] LHL : \lim_{x \to {\frac{π}{2}}^{-}} [\frac{2 x - π}{\cos x}] Let x = \frac{π}{2} - h If x \to \frac{π}{2}, then we have : h \to 0 = \lim_{h \to 0} [\frac{2 (\frac{π}{2} - h) - π}{\cos (\frac{π}{2} - h)}] = \lim_{h \to 0} [\frac{π - 2 h - π}{\sin h}] = - 2 RHL : \lim_{x \to {\frac{π}{2}}^{+}} [\frac{2 x - π}{\cos x}] Let x = \frac{π}{2} + h If x \to \frac{π}{2}, then we have : h \to 0 = \lim_{h \to 0} [\frac{2 (\frac{π}{2} + h) - π}{\cos (\frac{π}{2} + h)}] = \lim_{h \to 0} [\frac{2 h}{- \sin h}] = - 2 \Rightarrow \lim_{x \to \frac{π}{2}} (\frac{2 x - π}{\cos x}) = - 2$

Page No 29.83:

Question 15:

Write the value of $\lim_{n \to \infty} \frac{1 + 2 + 3 + . . . + n}{n^{2}} .$

Answer:

$\lim_{n \to \infty} (\frac{1 + 2 + 3 . . . n}{n^{2}}) = \lim_{n \to \infty} (\frac{n (n + 1)}{2 n^{2}}) = \lim_{n \to \infty} (\frac{(n + 1)}{2 n}) = \frac{1}{2} \lim_{n \to \infty} (1 + \frac{1}{n}) = \frac{1}{2}$

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