Contents
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 αm.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 hसे 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 :
[$\sin A=2\sin\frac{x}{2} \cos \frac{x}{2}$
$\cos x=\cos^{2} \frac{x}{2}-\sin^2 \frac{x}{2}$]
$=\lim_{x\rightarrow \pi}\dfrac{\cos ^{2} \frac{x}{4}+\sin ^{2} \frac{x}{4}-2 \sin \frac{x}{4} \cdot \cos \frac{x}{4}}{\left(\cos ^{2} \frac{x}{4}-\sin ^{2} \frac{x}{4}\right)\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)}$
$=\lim_{x\rightarrow \pi}\dfrac{\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)^{2}}{\left(\cos \frac{x}{4}+\sin \frac{x}{4}\right)\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)}$
$=\frac{1}{\cos \frac{\pi}{4}+\sin \frac{\pi}{4}}$
$=\frac{1}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt 2}}$
$=\frac{1}{\frac{2}{\sqrt{2}}}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$