KC Sinha Solution Class 11 Chapter 24 Limit (सीमा) Exercise 24.2

 Exercise 24.2

Question 1

निम्नलिखित सीमाओ को ज्ञात कीजिए
(Evaluate the following limits)
(i) $\lim_{\theta \rightarrow 0}\frac{\sin 5\theta}{\tan 6\theta}$Sol :
अंश तथा हर मे θ से भाग देने पर

$=\lim_{\theta\rightarrow 0}\dfrac{\frac{\sin 5\theta}{5\theta}\times 5}{\frac{\tan 6\theta}{6\theta}\times 6}$

$=\frac{5}{6}\dfrac{\lim_{\theta \rightarrow 0}\frac{\sin 5\theta}{5\theta}}{\lim_{\theta \rightarrow 0}\frac{\tan 6\theta}{6\theta }}$

$=\frac{5}{6}\times \frac{1}{1}=\frac{5}{6}$


(ii) $\lim_{x \rightarrow 0}\frac{\sin 2x}{\sin 3x}$
Sol :
$=\lim_{x\rightarrow 0}\frac{\sin 2x}{\sin 3x}$

अंश तथा हर मे x से भाग देने पर

$=\lim_{x\rightarrow 0}\dfrac{\frac{\sin 2x}{2x}\times 2}{\frac{\sin 3x}{3x}\times 2}$

$=\frac{2}{3}\dfrac{\lim_{x\rightarrow 0}\frac{\sin 2x}{2x}}{\lim_{x\rightarrow 0}\frac{\sin 3x}{3x}}$

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

(vi) $\lim_{x\rightarrow 0}\frac{\sin ax}{bx}$
Sol :
$=\frac{1}{b}\lim_{x\rightarrow 0}\frac{\sin ax}{ax}\times a$

$=\frac{a}{b}\lim_{x\rightarrow 0}\frac{\sin ax}{ax}$

$=\frac{a}{b}(1)$

$=\frac{a}{b}$


(viii) $\lim_{x\rightarrow 0}x\sec x$
Sol :
=0.sec0
=0×1
=0

(ix) $\lim_{x\rightarrow 0}\frac{\cos x}{\pi -x}$
Sol :
$=\dfrac{\lim_{x\rightarrow 0}\cos x}{\lim_{x\rightarrow 0}(\pi -x)}$

$=\frac{1}{\pi-0}=\frac{1}{\pi}$


(x) $\lim_{x\rightarrow 0}\frac{\sin ax+bx}{ax+\sin bx}$,a,b,a+b≠0
Sol :
अंश तथा हर मे x से भाग देने पर

$=\lim_{x\rightarrow 0}\dfrac{\frac{\sin ax}{x}+\frac{bx}{x}}{\frac{ax}{x}+\frac{\sin bx}{x}}$

$=\dfrac{\lim_{x\rightarrow 0}\frac{\sin ax}{ax}\times a+\lim_{x\rightarrow 0}b}{\lim_{x\rightarrow 0}a+\lim_{x\rightarrow 0}\frac{\sin bx}{bx}\times b}$

$=\dfrac{a\lim_{x\rightarrow 0}\frac{\sin ax}{ax}+b}{a+\lim_{x\rightarrow 0}\frac{\sin bx}{bx}}$

$=\frac{a(1)+b}{a+b(1)}$

$=\frac{a+b}{a+b}$

=1

(xi) $\lim_{x\rightarrow 0}\frac{ax+x\cos x}{b\sin x}$
Sol :
अंश तथा हर मे x से भाग देने पर

$=\lim_{x\rightarrow 0}\dfrac{\frac{ax}{x}+\frac{x\cos x}{x}}{\frac{b\sin x}{x}}$

$=\dfrac{\lim_{x\rightarrow 0}a+\lim_{x\rightarrow 0}\cos x}{b\lim_{x\rightarrow 0}\frac{\sin x}{x}}$

$=\frac{a+1}{b(1)}$

$=\frac{a+1}{b}$

Question 2

(i) निकाले(Find)
$\lim_{\theta \rightarrow 0}\theta (3\text{cosec} 2\theta -2\cot 3\theta)$
Sol :
$\lim_{\theta \rightarrow 0}\theta (3\operatorname{cosec} 2\theta -2\cot 3\theta)$

$=\lim_{\theta \rightarrow 0}\theta \left(\frac{3}{\sin 2\theta}-\frac{2}{\tan 3\theta}\right)$

$=\lim_{\theta \rightarrow 0}\left(\frac{3\theta}{\sin 2\theta}-\frac{2\theta}{\tan 3\theta}\right)$

$=\lim_{\theta \rightarrow 0}\left(\frac{3}{\frac{\sin 2\theta}{\theta}}-\frac{2}{\frac{\tan 3\theta}{\theta}}\right)$

$=\left(\dfrac{3}{\lim_{\theta \rightarrow 0}\frac{\sin 2\theta}{2\theta}\times 2}-\dfrac{2}{\lim_{\theta \rightarrow 0}\frac{\tan 3\theta}{3\theta}\times 3}\right)$

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

$=\frac{9-4}{6}=\frac{5}{6}$

(ii) निकाले(Find) $\lim_{x\rightarrow 0}\frac{\sin 7x-\sin x}{\sin 6x}$
Sol :
अंश तथा हर मे x से भाग देने पर

$=\lim_{x\rightarrow 0}\dfrac{\frac{\sin 7x}{x}-\frac{\sin x}{x}}{\frac{\sin 6x}{x}}$

$=\dfrac{\lim_{x\rightarrow 0}\frac{\sin 7x}{x}\times 7-\lim_{x\rightarrow 0}\frac{\sin x}{x}}{\lim_{x\rightarrow 0} \frac{\sin 6x}{x}\times 6}$

$=\frac{1(7)-1}{1\times 6}=\frac{6}{6}$

=1

Question 3

$\lim_{x\rightarrow 0}(\operatorname{cosec} x-\cot x)$
Sol :
$=\lim_{x\rightarrow 0}\left(\frac{1}{\sin x}-\frac{\cos x}{\sin x}\right)$

$=\lim_{x\rightarrow 0}\left(\frac{1-\cos x}{\sin x}\right)$

$=\lim_{x\rightarrow 0}\frac{2\sin^2 \frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}$

$=\lim_{x\rightarrow 0}\tan \frac{x}{2}$

$=\frac{\tan 0}{2}$

=tan 0=0

Question 4

निकाले(Find) $\lim_{x\rightarrow 0}\frac{\cos 7x-\cos 9x}{\cos 3x-\cos 5x}$

Sol :

$\left[\cos C-\cos D=2\sin \frac{C+D}{2} \sin\frac{D-C}{2}-2\sin \frac{C+D}{2}\sin \frac{C-D}{2}\right]$

$=\lim_{x\rightarrow 0}\dfrac{2\sin \frac{7x+9x}{2}\sin \frac{9x-7x}{2}}{2\sin \frac{3x+5x}{2}\sin \frac{5x-3x}{2}}$

$=\lim_{x\rightarrow 0}\frac{\sin 8x \sin x}{\sin 4x \sin x}$

अंश तथा हर मे x से भाग देने पर

$=\lim_{x\rightarrow 0}\dfrac{\frac{\sin 8x}{x}}{\frac{\sin 4x}{x}}$

$=\dfrac{\lim_{x\rightarrow 0}\frac{\sin 8x}{8x}\times 8}{\lim_{x\rightarrow 0} \frac{\sin 4x}{4x}\times 4}$

$=\frac{8}{4}=2$

Question 5

निकाले (Find) $\lim_{x\rightarrow 0}\frac{3\sin x-\sin 3x}{x^3}$

Sol :

[sin3x=3sinx-4sin³x

4sin³x=3sinx-sin3x]

$=\lim_{x\rightarrow 0}\frac{4\sin ^3 x}{x^3}$

$=4\lim_{x\rightarrow 0}\left(\frac{\sin x}{x}\right)^3$

=4(1)³

=4(1)

=4

Question 6

निकाले (Find) $\lim_{x\rightarrow 0}\frac{1-\cos kx}{x^2}$,(k≠0)

Sol :

$\left[1-\cos 2x=2\sin^2 x\\1-\cos x=2\sin^2 \frac{x}{2}\\1-\cos kx=2\sin^2 \frac{kx}{2}\right]$

$=\lim_{x\rightarrow 0}\dfrac{2\sin^2 \frac{kx}{2}}{x^2}$

$=2\lim_{x\rightarrow 0}\dfrac{\sin ^2 \frac{kx}{2}}{\frac{k^2}{4}x^2}\times \frac{k^2}{4}$

$=2\frac{k^2}{4}\lim_{x\rightarrow 0}\left(\dfrac{\sin \frac{kx}{2}}{\frac{kx}{2}}\right)^2$

$=\frac{k^2}{2}\times (1)^2$

$=\frac{k^2}{2}$

Question 7

निकाले (Find) $\lim_{x\rightarrow 0}\frac{1-\cos x}{x^2}$

Sol :

$=\lim_{x\rightarrow 0}\frac{2\sin^2 \frac{x}{2}}{x^2}$

$=2\lim_{x\rightarrow 0}\dfrac{\sin^2 \frac{x}{2}}{\frac{x^2}{4}}\times \frac{1}{4}$

$=\frac{2}{4}\lim_{x\rightarrow 0}\left(\dfrac{\sin^2 \frac{x}{2}}{\frac{x^2}{4}}\right)^2$

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

Question 8

निकाले (Find)

(i) $\lim_{x\rightarrow 0}\frac{1-\cos 4x}{1-\cos 5x}$

Sol :

$=\lim_{x\rightarrow 0}\frac{1-\cos 4x}{1-\cos 5x}$

$=\lim_{x\rightarrow 0}\frac{2\sin^2 2x}{2\sin ^2 \frac{5x}{2}}$

अंश तथा हर मे x² से भाग देने पर

$=\lim_{x\rightarrow 0}\dfrac{\frac{\sin ^2 2x }{x^2}}{\frac{\sin ^2 \frac{5x}{2}}{x^2}}$

$=\dfrac{\lim_{x\rightarrow 0}\frac{\sin^2 2x}{4x^2}\times 4}{\lim_{x\rightarrow 0}\dfrac{\sin^2 \frac{5x}{2}}{\frac{25}{4}x^2}\times \frac{25}{4}}$

$=\dfrac{4}{\frac{25}{4}}\dfrac{\lim_{x\rightarrow 0}\left(\frac{\sin 2x}{2x}\right)^2}{\lim_{x\rightarrow 0}\left(\frac{\sin \frac{5x}{2}}{\frac{5}{2}x}\right)^2}$

$=\frac{16}{25}\times \frac{(1)^2}{(1)^2}$

$=\frac{16}{25}$

(ii) $\lim_{x\rightarrow 0}\frac{\cos 2x-1}{\cos x-1}$

Sol :

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

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

$=\lim_{x\rightarrow 0}\frac{2\sin ^2 x}{2\sin ^2 \frac{x}{2}}$

अंश तथा हर मे x² से भाग देने पर

$=\lim_{x\rightarrow 0}\dfrac{\frac{\sin ^2 x}{x^2}}{\frac{\sin ^2 \frac{x}{2}}{x^2}}$

$=\dfrac{\lim_{x\rightarrow 0}\frac{\sin ^2 x}{x^2}}{\lim_{x\rightarrow 0}\dfrac{\sin^2 \frac{x}{2}}{\frac{x^2}{4}}\times \frac{1}{4}}$

$=4\dfrac{\lim_{x\rightarrow 0}\left(\frac{\sin x}{x}\right)^2}{\lim_{x\rightarrow 0}\left(\dfrac{\sin \frac{x}{2}}{\frac{3}{2}}\right)^2}$

$=\frac{4(1)^2}{(1)^2}$

=4

Question 9

निकाले (Find) $\lim_{x\rightarrow 0}\frac{1-\cos mx}{1-\cos nx}$ जहाँ m और n स्थिर अशून्य वास्तविक संख्याएँ है।(where m and n are fixed non zero real number)

Sol :

$=\lim_{x\rightarrow 0}\frac{1-\cos mx}{1-\cos nx}$

$=\lim_{x\rightarrow 0}\frac{2\sin^2 \frac{mx}{2}}{2\sin^2 \frac{nx}{2}}$

अंश तथा हर मे x² से भाग देने पर

$=\lim_{x\rightarrow 0}\dfrac{\frac{\sin^2 \frac{mx}{2}}{x^2}}{\frac{\sin^2 \frac{nx}{2}}{x^2}}$

$=\lim_{x\rightarrow 0}\dfrac{\frac{\sin^2 \frac{mx}{2}}{\frac{m^2x^2}{4}}\times \frac{m^2 }{4}}{\frac{\sin^2 \frac{nx}{2}}{\frac{n^2}{4}x^2}\times \frac{n^2}{4}}$

$=\frac{m^2}{n^2}\dfrac{\lim_{x\rightarrow 0}\left(\dfrac{\sin \frac{mx}{2}}{\frac{mx}{2}}\right)^2}{\lim_{x\rightarrow 0}\left(\dfrac{\sin \frac{nx}{2}}{\frac{nx}{2}}\right)^2}$

$=\frac{m^2}{n^2}\times \frac{(1)^2}{(1)^2}$

$=\frac{m^2}{n^2}$

Question 10

निकाले (Find) $\lim_{x\rightarrow 0}\frac{\tan x-\sin x}{\sin^3 x}$

Sol :

$=\lim_{x\rightarrow 0}\dfrac{\frac{\sin x}{\cos x}-\sin x}{\sin ^3 x}$

$=\lim_{x\rightarrow 0}\dfrac{\sin x\left(\frac{1}{\cos x}-1\right)}{\sin ^3 x^2}$

$=\lim_{x\rightarrow 0}\dfrac{\frac{1-\cos x}{\cos x}}{\frac{\sin ^2 x}{1}}$

$=\lim_{x\rightarrow 0}\frac{1-\cos x}{\cos x.\sin^2 x}$

$=\lim_{x\rightarrow 0}\frac{2\sin ^2 \frac{x}{2}}{\cos x.\sin^2 x}$

अंश तथा हर मे x² से भाग देने पर

$=2\lim_{x\rightarrow 0}\dfrac{\frac{\sin^2 \frac{x}{2}}{x^2}}{\frac{\cos x.\sin^2 x}{x^2}}$

$=2\dfrac{\lim_{x\rightarrow 0}\dfrac{\sin^2 \frac{x}{2}}{\frac{x^2}{4}}\times \frac{1}{4}}{\lim_{x\rightarrow 0}\cos x.\lim_{x\rightarrow 0} \frac{\sin ^2 x}{x^2}}$

$=\frac{2}{4}\dfrac{\lim{x\rightarrow 0}\left(\dfrac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2}{\lim_{x\rightarrow 0}\cos x \lim_{x\rightarrow 0}\left(\frac{\sin x}{x}\right)^2}$

$=\frac{1}{2}\frac{(1)^2}{1\times (1)^2}$

$=\frac{1}{2}$

Question 11

निकाले (Find) $\lim_{x\rightarrow 0}\frac{2\sin x-\sin 2x}{x^3}$

Sol :

$=\lim_{x\rightarrow 0}\frac{2\sin x-\sin x\cos x}{x^3}$

$=\lim_{x\rightarrow 0}\frac{2\sin x(1-\cos x)}{x^3}$

$=\lim_{x\rightarrow 0}\frac{2\sin x.2\sin^2 \frac{x}{2}}{x^3}$

$=\lim_{x\rightarrow 0}\frac{2\sin x.\sin^2 \frac{x}{2}}{\frac{x^3}{4}}$

$=\lim_{x\rightarrow 0}\frac{2\sin x.\sin^2 \frac{x}{2}}{x.\frac{x^2}{4}}$

$=\lim_{x\rightarrow 0} \frac{\sin x}{x}\lim_{x\rightarrow 0} \left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2$

=1×(1)²

=1

Question 12

निकाले (Find)$\lim_{x\rightarrow 0}\frac{\sin (2+x)-\sin (2-x)}{x}$

Sol :

[sin(A+B)-sin(A-B)=2cosAsinB]

$=\lim_{x\rightarrow 0} \frac{2\cos 2.\sin x}{x}$

$=2 \cos 2 \lim_{x\rightarrow 0} \frac{\sin x}{x}$

=2cos2×1

=2cos2

Question 13

निकाले (Find)$\lim_{x\rightarrow 0}\frac{\tan x-\sin x}{x^2}$

Sol :

$=\lim_{x\rightarrow 0}\frac{\frac{\sin x}{\cos x}-\sin x}{x^2}$

$=\lim_{x\rightarrow 0}\frac{\sin x \left(\frac{1}{\cos x}-1\right)}{x^2}$

$=\lim_{x\rightarrow 0}\frac{\sin x \left(\frac{1-\cos x}{\cos x}\right)}{x^2}$

$=\lim_{x\rightarrow 0} \frac{\sin x.2\sin ^2 \frac{x}{2}}{\cos x.x^2}$

$=2\lim_{x\rightarrow 0} \frac{\sin x. \sin^2 \frac{x}{2}}{\cos x.\frac{x^2}{4}\times 4}$

$=\frac{2}{4} \dfrac{\lim_{x\rightarrow 0}\sin x}{\lim_{x\rightarrow 0} \cos x}\lim_{x\rightarrow 0} \left(\dfrac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2$

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

=0

Question 14

निकाले (Find) $\lim_{x\rightarrow 0}\frac{1-\sec x}{x^2}$

Sol :

$=\lim_{x\rightarrow 0}\frac{1-\frac{1}{\cos x}}{x^2}$

$=\lim_{x\rightarrow 0}\dfrac{\frac{\cos x-1}{\cos x}}{\frac{x^2}{1}}$

$=\lim_{x\rightarrow 0}\dfrac{\frac{\cos x-1}{\cos x}}{\frac{x^2}{1}}$

$=\lim_{x\rightarrow 0}\frac{1-\cos x}{\cos x.x^2}$

$=\lim_{x\rightarrow 0}\dfrac{\sin^2 \frac{x}{2}}{\cos x.x^2}$

$=2\lim_{x\rightarrow 0}\dfrac{\sin^2 \frac{x}{2}}{\cos x.\frac{x^2}{4}\times 4}$

$=\frac{-2}{4}\dfrac{\lim_{x\rightarrow 0}\left(\frac{\sin^2 \frac{x}{2}}{\frac{x}{2}}\right)^2}{\lim_{x\rightarrow 0}\cos x}$

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

Question 15

निकाले (Find) $\lim_{\theta \rightarrow 0}\frac{1-\cos x}{\sin^2 2\theta}$

Sol :

$=\lim_{\theta \rightarrow 0}\frac{2\sin^2 \frac{\theta }{2}}{(2\sin \theta .\cos \theta)^2}$

$=\lim_{\theta \rightarrow 0}\frac{2\sin^2 \frac{\theta }{2}}{4\sin^2 \theta .\cos^2 \theta}$

$=\lim_{\theta \rightarrow 0}\frac{\sin^2 \frac{\theta }{2}}{2\left(2\sin \frac{\theta}{2} .\cos \frac{\theta}{2}\right)\cos^2 \theta}$

$=\lim_{\theta \rightarrow 0}\frac{\sin^2 \frac{\theta }{2}}{2\times 4\sin^2 \frac{\theta}{2} .\cos^2 \frac{\theta}{2}.\cos^2 \theta}$

$=\frac{1}{8}\lim_{\theta \rightarrow 0}\dfrac{1}{\cos^2 \frac{\theta}{2}.\cos^2 \theta}$

$=\frac{1}{8}\dfrac{1}{\lim_{\theta \rightarrow 0}\cos^2 \frac{\theta}{2}.\lim_{\theta \rightarrow 0}\cos^2 \theta}$

$=\frac{1}{8}\times \frac{1}{\cos^2 0.\cos^2 0}$

$=\frac{1}{8}\times \frac{1}{(1)^2\times (1)^2}=\frac{1}{8}$

Question 16

निकाले (Find) $\lim_{\theta \rightarrow 0}\frac{\tan ^2 x-\sin ^2 x}{x^4}$

Sol :

$=\lim_{\theta \rightarrow 0}\dfrac{\frac{\sin ^2 x}{\cos ^2 x}-\sin ^2 x}{x^4}$

$=\lim_{\theta \rightarrow 0}\dfrac{\sin ^2 x \left(\frac{1}{\cos ^2 x}-1\right)}{x^4}$

$=\lim_{\theta \rightarrow 0}\dfrac{\sin ^2 x .\sin^2 x}{\cos^2 x.x^4}$

$=\lim_{\theta \rightarrow 0}\dfrac{\sin ^4 x}{\cos^2 x.x^4}$

$=\dfrac{\lim_{\theta \rightarrow 0}\left(\frac{\sin ^4 x}{x}\right)^4}{\lim_{\theta \rightarrow 0}\cos^2 x}$

$=\frac{(1)^4}{\cos^2 0}=\frac{1}{1^2}$

=1

Question 17

निकाले (Find) $\lim_{\theta \rightarrow 0}\frac{\tan 2 x-\sin 2 x}{x^3}$

Sol :

$=\lim_{\theta \rightarrow 0}\dfrac{\frac{\sin 2 x}{\cos 2 x}-\sin 2 x}{x^3}$

$=\lim_{\theta \rightarrow 0}\dfrac{\sin 2 x \left(\frac{1}{\cos 2 x}-1\right)}{x^3}$

$=\lim_{\theta \rightarrow 0}\dfrac{\sin 2 x \left(\frac{1-\cos 2 x}{\cos 2x}\right)}{x^3}$

$=\lim_{\theta \rightarrow 0}\dfrac{\sin 2x.2\sin^2 x}{\cos 2x.x^3}$

$=2\lim_{\theta \rightarrow 0}\dfrac{\frac{\sin 2x}{x}-\frac{\sin^2 x}{x^2}}{\cos 2x}$

$=\dfrac{2\lim_{\theta \rightarrow 0}\frac{\sin 2x}{2x}\times 2 \times \lim_{x\rightarrow 0}\left(\frac{\sin x}{x}\right)^2}{\lim_{x\rightarrow 0}\cos 2x}$

$=\dfrac{2\times 1\times 2\times (1)^2}{\cos 2(0)}$

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

Question 18

निकाले (Find) $\lim_{x\rightarrow 0}\frac{3\sin x-\sin x}{x(\cos 2x-\cos 4x)}$

Sol :

[sin3x=3sinx-4sin³x

4sin³x=3sinx-sin3x

cosC-cosD=2sin$\frac{C+D}{2}\sin \frac{D-C}{2}$]

$=\lim_{x\rightarrow 0} \frac{4\sin ^3 x}{x.2\sin \frac{2x+4x}{2}.\sin \frac{4x-2x}{2}}$

$=\lim_{x\rightarrow 0} \frac{\sin ^3 x}{x.\sin 3x.\sin x}$

अंश तथा हर मे x² से भाग देने पर

$=\lim_{x\rightarrow 0} \dfrac{\frac{\sin ^3 x}{x^2}}{x.\frac{\sin 3x.\sin x}{x^2}}$

$=\lim_{x\rightarrow 0} \dfrac{\frac{\sin ^3 x}{x^3}}{x.\frac{\sin 3x}{x}.\frac{\sin x}{x}}$

$=\dfrac{2\lim_{x\rightarrow 0}\left(\frac{\sin x}{x}\right)^3}{\lim_{x\rightarrow 0}\frac{\sin 3x}{3x}\times 3.\lim_{x\rightarrow 0}\frac{\sin x}{x}}$

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

Question 19

निकाले (Find) $\lim_{x\rightarrow 0}\frac{\cos ax-\cos bx}{\cos x-1}$

Sol :

[cos C-cos D$=-2\sin \frac{C+D}{2}\sin \frac{C-D}{2}$]

$=\lim_{x\rightarrow 0}\dfrac{-2\sin \frac{ax+bx}{2}\sin \frac{ax-bx}{2}}{-(1-\cos x)}$

$=\lim_{x\rightarrow 0}\dfrac{2\sin \left(\frac{a+b}{2}\right) x \sin \left(\frac{a-b}{2}\right)x}{2\sin^2 \frac{x}{2}}$

अंश तथा हर मे x² से भाग देने पर

$=\lim_{x\rightarrow 0}\dfrac{\frac{\sin\left(\frac{a+b}{2}\right)x.\sin \left(\frac{a-b}{2}\right)x}{x^2}}{\frac{\sin^2 \frac{x}{2}}{x^2}}$

$=\lim_{x\rightarrow 0}\dfrac{\frac{\sin\left(\frac{a+b}{2}\right)x}{\left(\frac{a+b}{2}\right)x}\times \left(\frac{a+b}{2}\right).\frac{\sin \left(\frac{a-b}{2}\right)x}{\left(\frac{a-b}{2}\right)x}\times \left(\frac{a-b}{2}\right)}{\frac{\sin^2 \frac{x}{2}}{\frac{x^2}{4}}\times \frac{1}{4}}$

$=4\left(\frac{a+b}{2}\right)\left(\frac{a-b}{2}\right).\dfrac{\lim_{x\rightarrow 0}\dfrac{\sin \left(\frac{a+b}{2}\right)x}{\left(\frac{a+b}{2}\right)x}\lim_{x\rightarrow 0}\dfrac{\sin \left(\frac{a-b}{2}\right)x}{\left(\frac{a-b}{2}\right)x}}{\lim_{x\rightarrow 0}\left(\dfrac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2}$

$=4\left(\frac{a^2-b^2}{4}\right)\frac{1\times 1}{(1)^2}$

=a²-b²

Question 20

निकाले (Find) $\lim_{\alpha \rightarrow 0}\frac{\sin \alpha^n}{(\sin \alpha)^m}$,n>m>0

Sol :

अंश तथा हर मे α^m.αⁿ से भाग देने पर

$=\lim_{\alpha \rightarrow 0}\dfrac{\frac{\sin \alpha ^n}{\alpha ^m.\alpha^n}}{\frac{(\sin \alpha)^m}{\alpha^m.\alpha^n}}$

$=\dfrac{\lim_{\alpha \rightarrow 0}\frac{\sin \alpha ^n}{\alpha ^n} \frac{1}{\alpha^m}}{\lim_{\alpha \rightarrow 0}\frac{(\sin \alpha)^m}{\alpha^m}\times \frac{1}{\alpha ^n}}$

$=\dfrac{\lim_{\alpha \rightarrow 0}\frac{\sin \alpha^n}{\alpha^n}\lim_{\alpha \rightarrow 0}\alpha^{n-m}}{\lim_{\alpha \rightarrow 0}\left(\frac{\sin \alpha}{\alpha}\right)^m}$

$=\frac{1\times 0}{(1)^m}=0$

Question 21

निकाले (Find) $\lim_{h\rightarrow 0}\dfrac{\sin(a+3h)-\sin (a+2h)+3\sin (a+h)-\sin a}{h^3}$

Sol :

$=\lim_{h\rightarrow 0}\dfrac{\left[\sin(a+3b)-\sin \alpha \right]-3\left[\sin(a+zh)-\sin(a+h)\right]}{h^3}$

$=\lim_{h\rightarrow 0}\dfrac{2\cos \frac{2a+3h}{2}\sin \frac{3h}{2}-3\times 2\cos \frac{2a+3h}{2}\sin \frac{h}{2}}{h^3}$

$\lim_{h\rightarrow 0}\dfrac{2\cos \frac{2a+3h}{2}\left[\sin \frac{3h}{2}-3\sin \frac{h}{2}\right]}{h^3}$

[sin3θ=3sin3θ-4sin³θ]

$\lim_{h\rightarrow 0}\dfrac{2\cos \frac{2a+3h}{2}\left[3\sin \frac{h}{2}-4\sin^3 \frac{h}{2}-3\sin \frac{h}{2}\right]}{h^3}$

$=-\lim_{h\rightarrow 0}\frac{\cos 2a+3h}{2}\times \frac{8\sin^3 \frac{h}{2}}{h^3}$

$=-\lim_{h\rightarrow 0}\frac{\cos 2a+3h}{2}\times \lim_{h\rightarrow 0}\left(\dfrac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^3$

$=\frac{-\cos 2a+3(0)}{2}\times (1)^3$

$=-\cos \frac{2a}{2}$

=-cos a

Question 22

निकालो (Find) $\lim_{h\rightarrow 0}\frac{\tan (a+2h)-2\tan (a+h)+\tan a}{h^2}$

Sol :

By L hopital's rule

$=\lim_{h\rightarrow 0}\frac{2\sec^2 (a+2h)-2\sec^2(a+h)}{2h}$

$=\lim_{h\rightarrow 0}\frac{2[\sec^2 (a+2h)-\sec^2(a+h)]}{2h}$

Again differentiating numerator(अंश) and denominator(हर)

$=\lim_{h\rightarrow 0}$2.2sec(a+2h).sec(a+2h).tan(a+2h)-sec(a+h).sec(a+h).tan(a+h)

$=\lim_{h\rightarrow 0}$4sec³(a+2h).tan(a+2h)-sec²(a+h).tan(a+h)

=4sec²a.tan a-2sec²a.tan a

=2sec²a.tan a

Question 23

निकालो (Find) 

(i) $\lim_{x\rightarrow \frac{\pi}{2}}\left(\frac{\pi}{2}-x \right)\tan x$

Sol :

माना $x=\frac{\pi}{2}+h$ ,जब $x\rightarrow \frac{\pi}{2}, h\rightarrow 0$

$=\lim_{h\rightarrow 0}\left[\frac{\pi}{2}-\left(\frac{\pi}{2}+h\right)\right].\tan \left(\frac{\pi}{2}+h\right)$

$=\lim_{h\rightarrow 0}\left[\frac{\pi}{2}-\frac{\pi}{2}-h\right](-\cot h)$

$=\lim_{h\rightarrow 0}\frac{-h}{-\tan h}$

=1

(ii) $\lim_{x\rightarrow \frac{\pi}{2}}\frac{\tan 2x}{x-\frac{x}{2}}$

Sol :

माना $x=\frac{\pi}{2}+h$ ,जब $x\rightarrow \frac{\pi}{2}, h\rightarrow 0$

$=\lim_{h\rightarrow 0}\frac{\tan 2\left(\frac{\pi}{2}+h\right)}{\frac{\pi}{2}+h-\frac{\pi}{2}}$

$=\lim_{h\rightarrow 0}\frac{\tan \left(\pi+2h\right)}{h}$

$\lim_{h\rightarrow 0}\frac{\tan 2h}{2h}\times 2$

=1×2

=2

Question 24

निकालो (Find) 

$\lim_{x\rightarrow \frac{\pi}{2}}\frac{1-\sin x}{\cos x}$

Sol :

माना $x=\frac{\pi}{2}-h$ ,जब $x\rightarrow \frac{\pi}{2}, h\rightarrow 0$

$=\lim _{h \rightarrow 0} \frac{1-\sin \left(\frac{\pi}{2}-4\right)}{\cos \left(\frac{\pi}{2}-h\right)}$

$=\lim _{h \rightarrow 0} \frac{-\cos h}{\sin h}$

$=\lim _{h \rightarrow 0} \frac{2 \sin^2 \frac{h}{2}}{2 \sin \frac{h}{2} \cos \frac{h}{2}}$

$=\lim _{h \rightarrow 0} \tan \frac{h}{2}=\tan 0$

=0

Question 25

निकालो (Find) 

(i) $\lim _{x \rightarrow \frac{\pi}{2}} \frac{2 \cos x}{\pi-2 x}$

Sol :

माना $x=\frac{\pi}{2}+h$ ,जब $x\rightarrow \frac{\pi}{2}, h\rightarrow 0$

$=\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{\pi}{2}+h\right)}{\pi-2\left(\frac{\pi}{2}+4\right)}$

$=\lim _{h \rightarrow 0} \frac{-2 \sin h}{\pi-\pi-2 h}$

$=\lim _{h \rightarrow 0} \frac{-2 \sin h}{-2 h}$

=1

(ii) $\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}$

Sol :

माना x=π+h , जब x⟶π, h⟶0

$=\lim _{h \rightarrow 0} \frac{\sin [\pi-(\pi+4)]}{\pi[\pi-(\pi+h)]}$

$=\lim _{h \rightarrow 0} \frac{\sin (\pi-\pi-h)}{\pi(\pi-\pi-h)}$

$=\frac{1}{\pi} \lim _{h \rightarrow 0} \frac{-\sin h }{- h}$

$=\frac{1}{\pi}(1)=\frac{1}{\pi}$

Question 26

निकालो (Find) 

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\left(\frac{\pi}{2}-x\right) \cot x}$

Sol :

माना $x=\frac{\pi}{2}+h$, जब $x \rightarrow \frac{\pi}{2}$ , h⟶0

$=\lim _{h \rightarrow 0} \frac{1-\sin \left(\frac{\pi}{2}+h\right)}{\left[\frac{\pi}{2}-\left(\frac{\pi}{2}+h\right)\right] \cdot \cot \left(\frac{\pi}{2}+h\right)}$

$=\lim _{h \rightarrow 0} \frac{1-\cos h}{\left[\frac{\pi}{2}-\frac{\pi}{2}-4\right]\left(-\tan h\right)}$

$=\lim _{h \rightarrow 0} \frac{1-\cos h}{h \cdot \tan h}$

$=\lim _{h \rightarrow 0} \frac{2 \sin ^{2} \frac{h}{2}}{h \cdot \tan h}$

अंश तथा हर मे h² से भाग करने पर,

$=2 \lim _{h \rightarrow 0} \frac{\frac{\sin ^{2} \frac{h}{2}}{4^{2}}}{\frac{h.\tan h}{h^{2}}}$

$=2 \lim _{h \rightarrow 0} \frac{\frac{\sin ^{2} \frac{h}{2}}{\frac{h^{2}}{4}} \times \frac{1}{4}}{\lim _{h \rightarrow 0} \frac{\tan h}{h}}$

$=\frac{2}{4} \frac{\lim _{h \rightarrow 0}\left(\frac{\sin \frac{h}{h}}{\frac{h}{2}}\right)^{2}}{\lim _{h \rightarrow 0} \frac{\tan h}{h}}$

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

$=\frac{1}{2}$

Question 27

निकालो (Find) 

$\lim _{x \rightarrow 1}(1-x) \tan \frac{\pi x}{2}$

Sol :

माना x=1+h , जब x⟶1, h⟶0

$=\lim _{h \rightarrow 0}[1-(1+h)] \tan \frac{\pi}{2}(1+h)$

$=\lim _{h \rightarrow 0}(1-1-h) \cdot \tan \left(\frac{\pi}{2}+\frac{\pi}{2} h\right)$

$=\lim _{h \rightarrow 0}-h\left(-\cot \frac{\pi}{2} h\right)$

$=\lim _{h \rightarrow 0} \frac{\frac{\pi}{2} h}{\tan \frac{\pi}{2} h} \times \frac{2}{\pi}$

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

Question 28

निकालो (Find) 

$\lim _{x \rightarrow \pi} \frac{1+\cos x}{\tan ^{2} x}$

Sol :

माान x=π+h , जब x⟶π, h⟶0

$=\lim _{h \rightarrow 0} \frac{1+\cos (\pi+h)}{\tan ^{2}(\pi+h)}$

$=\lim _{h \rightarrow 0} \frac{1-\cos h}{\tan ^{2} h}$

$=\lim _{h \rightarrow 0} \frac{2 \sin ^{2} \frac{h}{2}}{\tan ^{2} h}$

$=2 \lim _{h \rightarrow 0} \frac{\sin ^{2} \frac{h}{2}}{\tan ^{2} h}$

अंश तथा हर मे h² से भाग देने पर

$=2 \lim _{h \rightarrow 0} \dfrac{\frac{\sin ^{2} \frac{h}{2}}{h^2}}{\frac{\tan ^{2}}{h^{2} }}$

$=2 \frac{\lim _{h \rightarrow 0} \frac{\sin ^{2} \frac{h}{2}}{\frac{h^{2}}{4}} \times \frac{1}{4}}{\lim_{h \rightarrow 0} \frac{\tan ^{2} h}{h^{2}}}$

$=\frac{2}{4} \cdot \frac{\lim _{h \rightarrow 0}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^{2}}{\lim _{h \rightarrow 0}\left(\frac{\tan h}{h}\right)^{2}}$

$=\frac{1}{2} \times \frac{(1)^{2}}{(1)^{2}}=\frac{1}{2}$

Question 29

निकालो (Find) 

$\lim _{x \rightarrow 1} \frac{\cos \pi x / 2}{1-\sqrt{x}}$

माना x=1+h , जब x⟶1, h⟶0

$=\lim _{h \rightarrow 0} \frac{\cos \frac{\pi}{2}(1+h)}{1-\sqrt{1+h}}$

$=\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{2}+\frac{\pi}{2} h\right)}{1-\sqrt{1+h}}$

$=\lim _{h \rightarrow 0} \frac{-\operatorname{sin} \frac{\pi}{2} h}{1-\sqrt{1+h}}$

$=\lim _{h \rightarrow 0} \frac{\sin \frac{\pi}{2} h}{\sqrt{1+h}-1}$

अंश तथा हर मे h से भाग देने पर

$=\lim _{h \rightarrow 0} \frac{\frac{\sin \frac{\pi}{2} h}{h}}{\frac{\sqrt{1+h}-1}{h}}$

$=\dfrac{\lim _{h \rightarrow 0} \frac{\sin \frac{\pi}{2} h}{\frac{\pi}{2} h} \times \frac{\pi}{2}}{\lim_{h\rightarrow 0} \frac{(1+h)^{\frac{1}{2}}-1^{\frac{1}{2}}}{(1+h)-1}}$

$=\frac{\frac{\pi}{2}}{\frac{1}{2} \cdot(1)^{\frac{1}{2}-1}}$

$=\frac{\pi}{1}=\pi$

Question 30

निकालो (Find) 

$\lim _{\theta \rightarrow \frac{\pi}{4}} \frac{1-\tan \theta}{1-\sqrt{2} \sin \theta}$

Sol :

By L Hospital rule

$=\lim _{\theta \rightarrow \frac{\pi}{4}} \frac{- \sec ^{2} \theta}{- \sqrt{2} \cos \theta}$

$=\frac{1}{\sqrt{2}} \lim_{\theta \rightarrow \frac{\pi}{4}} \sec ^{3} \theta$

$=\frac{1}{\sqrt{2}} \times \sec ^{3} \frac{\pi}{4}$

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

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

Question 31

निकालो (Find) 

$\lim _{y \rightarrow a} \sin \frac{y-a}{2} \tan \frac{\pi y}{2 a}$

Sol :

$=\lim _{y \rightarrow a} \frac{\sin \frac{y-a}{2}}{\cot \frac{\pi y}{2a}}$

By L Hospital Rule

$=\lim _{y \rightarrow a} \frac{\frac{1}{2} \cos \frac{y-a}{2}}{-\frac{\pi}{2 a} \operatorname{cosec}^{2} \frac{\pi y}{2 a}}$

$=-\frac{a}{\pi} \lim _{y \rightarrow a} \frac{\cos \frac{y-a}{2}}{\operatorname{cosec}^{2} \frac{\pi y}{2 a}}$

$=-\frac{a}{\pi} \cdot \frac{\cos \frac{a-a}{2}}{\operatorname{cosec}^{2} \frac{\pi a}{2 a}}$

$=-\frac{a}{\pi} \times \frac{1}{(1)^{2}}=\frac{-a}{\pi}$

Question 32

निकालो (Find) 

$\lim _{\theta \rightarrow \frac{\pi}{2}} \frac{\sec \theta-\tan \theta}{\pi-2 \theta}$

Sol :

माना $\theta=\frac{\pi}{2}-h$ , जहाँ $\theta \rightarrow \frac{\pi}{2}$ ,h⟶0

$=\lim _{h \rightarrow 0} \frac{\sec \left(\frac{\pi}{2}-h\right)-\tan \left(\frac{\pi}{2}-4\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}$

$=\lim _{h \rightarrow 0} \frac{\operatorname{cosec h}-\cot h}{\pi-\pi+2 h}$

$=\lim _{h \rightarrow 0} \frac{\frac{1}{\sinh }-\frac{\cos h}{\sin h}}{2 h}$

$=\lim _{h \rightarrow 0} \frac{\frac{1-\cos h}{\sin h}}{2 h}$

$=\lim _{h \rightarrow 0} \frac{\frac{2 \sin ^{2} \frac{h}{2}}{2 \sin \frac{h}{2} \cos \frac{h}{2}}}{2 h}$

$=\lim _{h \rightarrow 0} \frac{\tan \frac{h}{2}}{2 h}$

$=\lim _{h \rightarrow 0} \frac{\tan \frac{h}{2}}{2 \frac{h}{2} \times 2}$

$=\frac{1}{4} \lim _{h \rightarrow 0} \frac{\tan \frac{h}{2}}{\frac{h}{2}}$

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

Question 33

निकालो (Find) 

$\lim _{\theta \rightarrow \frac{\pi}{2}} \frac{\cot \theta-\cos \theta}{\cos ^{3} \theta}$

Sol :

माना $\theta=\frac{\pi}{2}-h$ , जहाँ $\theta \rightarrow \frac{\pi}{2}$ ,h⟶0

$=\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{2}-h\right)-\cos \left(\frac{\pi}{2}-h\right)}{\cos ^{3}\left(\frac{\pi}{2}-h\right)}$

$=\lim _{h \rightarrow 0} \frac{\tan h-\sin h}{\operatorname{sin}^{3} h}$

$=\lim _{h \rightarrow 0} \dfrac{\frac{\sin h}{\cos h}-\sin h}{\sinh ^{3} h}$

$=\lim _{h \rightarrow 0} \frac{\sin h \left(\frac{1}{\cos h}-1\right)}{\sin ^{3} h^{2}}$

$=\lim _{h \rightarrow 0}\dfrac{\frac{1-\cos h}{\cos h}}{\frac{\sin^2 h}{1}}$

$=\lim _{h \rightarrow 0} \frac{1-\cos h}{\cos h \cdot \sin ^{2} h}$

$=\lim _{h \rightarrow 0} \frac{2 \sin ^{2} \frac{h}{2}}{\cos h \cdot \sin ^{2} h}$

अंश तथा हर मे h² से भाग देने पर

$=\lim _{h \rightarrow 0} \dfrac{\frac{2 \sin ^{2} \frac{h}{2}}{h^2}}{\cos h \cdot \frac{\sin^{2} h}{h^{2}}}$

$=\dfrac{2 \lim _{h \rightarrow 0} \frac{\sin ^{2} \frac{h}{2}}{\frac{h^{2}}{4}} \times \frac{1}{4}}{\lim _{h \rightarrow 0} \cos h \cdot \lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right)^{2}}$

$=\frac{2}{4} \frac{\lim _{h \rightarrow 0}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^{2}}{\lim _{h \rightarrow 0} \cos h \cdot \lim _{h \rightarrow 0}\left(\frac{\operatorname{sin} h}{h}\right)^{2}}$

$=\frac{1}{2}\times \frac{(1)^{2}}{1 \times(1)^{2}}=\frac{1}{2}$

Question 33

निकालो (Find) 

$=\lim _{x \rightarrow 1} \frac{1+\cos \pi x}{\tan ^{2} \pi x}$

Sol :

$=\lim _{x \rightarrow 1} \frac{1+\cos \pi x}{\tan ^{2} \pi x}$

By L Hospital Rule

$=\lim _{x \rightarrow 1} \frac{-\pi \sin \pi x}{2 \tan \pi x.\sin \pi \cdot \sec ^{2} \pi x \cdot \pi}$

$=-\frac{1}{2} \lim _{x \rightarrow 1} \frac{\sin \pi x}{\frac{\sin \pi x}{\cos \pi x} \cdot \sec ^{2} \pi x}$

$=-\frac{1}{2} \lim _{x \rightarrow 1} \cos \pi x \cdot \cos ^{2} \pi x$

$=-\frac{1}{2} \lim _{x \rightarrow 1} \cos ^{3} \pi x$

$=-\frac{1}{2}\times \cos ^{3}\left(\pi \times 1\right)$

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

Question 35

निकालो (Find) 

$\lim _{x \rightarrow y} \frac{\sin x-\sin y}{x-y}$

Sol :

By L Hospital Rule

$=\lim _{x \rightarrow y} \frac{\cos x}{1}=\cos y$

Question 36

निकालो (Find) 

$\lim _{\alpha \rightarrow \frac{\pi}{4}} \frac{\sin \alpha-\cos \alpha}{\alpha-\frac{\pi}{4}}$

Sol :

By L Hospital Rule

$=\lim_{\alpha \rightarrow \frac{\pi}{4}} \frac{\cos \alpha+\sin \alpha}{1}$

$=\cos \frac{\pi}{4}+\sin \frac{\pi}{4}$

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

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

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

Question 37

निकालो (Find) 

$\lim _{y \rightarrow 0} \frac{y \cos \theta-\theta \cos y}{y-\theta}$

Sol :

By L Hospital Rule

$=\lim _{y \rightarrow 0} \frac{\cos \theta+\theta \sin y}{1}$

=cosθ+θsinθ 

Question 38

निकालो (Find) 

$\lim _{x \rightarrow a} \frac{\sin x-\sin a}{\sqrt{x}-\sqrt{a}}$

Sol :

By L Hospital Rule

$=\lim _{x \rightarrow a} \frac{\cos x}{\frac{1}{2 \sqrt{x}}}$

$=\lim _{x \rightarrow a} 2 \sqrt{x} \cos x$

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

Question 39

निकालो (Find) 

$\lim _{x \rightarrow \pi} \frac{1-\sin \frac{x}{2}}{\cos \frac{x}{2}\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)}$

Sol :

[$\sin A=2\sin\frac{x}{2} \cos \frac{x}{2}$

$\cos x=\cos^{2} \frac{x}{2}-\sin^2 \frac{x}{2}$]

$=\lim_{x\rightarrow \pi}\dfrac{\cos ^{2} \frac{x}{4}+\sin ^{2} \frac{x}{4}-2 \sin \frac{x}{4} \cdot \cos \frac{x}{4}}{\left(\cos ^{2} \frac{x}{4}-\sin ^{2} \frac{x}{4}\right)\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)}$

$=\lim_{x\rightarrow \pi}\dfrac{\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)^{2}}{\left(\cos \frac{x}{4}+\sin \frac{x}{4}\right)\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)}$

$=\frac{1}{\cos \frac{\pi}{4}+\sin \frac{\pi}{4}}$

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

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