lim _{x rightarrow frac{pi}{2}} frac{1-tan fr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $L = \lim _{x ightarrow \frac{\pi}{2}} 	an(\frac{\pi}{4} - \frac{x}{2}) \cdot \frac{1-\sin x}{(\pi-2 x)^3}$.Substitute $x = \frac{\pi}{2} +

Vedclass · Accepted Answer

$\frac{1}{32}$

Vedclass · Answer

(A) Let $L = \lim _{x ightarrow \frac{\pi}{2}} 	an(\frac{\pi}{4} - \frac{x}{2}) \cdot \frac{1-\sin x}{(\pi-2 x)^3}$.Substitute $x = \frac{\pi}{2} + h$,where $h ightarrow 0$ as $x ightarrow \frac{\pi}{2}$.Then $\frac{\pi}{4} - \frac{x}{2} = \frac{\pi}{4} - \frac{\pi}{4} - \frac{h}{2} = -\frac{h}{2}$.Also,$\pi - 2x = \pi - 2(\frac{\pi}{2} + h) = -2h$.$1 - \sin x = 1 - \sin(\frac{\pi}{2} + h) = 1 - \cos h = 2\sin^2(\frac{h}{2})$.Substituting these into the limit:$L = \lim _{h ightarrow 0} 	an(-\frac{h}{2}) \cdot \frac{2\sin^2(\frac{h}{2})}{(-2h)^3}$.$L = \lim _{h ightarrow 0} -	an(\frac{h}{2}) \cdot \frac{2\sin^2(\frac{h}{2})}{-8h^3}$.$L = \lim _{h ightarrow 0} \frac{	an(\frac{h}{2})}{h} \cdot \frac{2\sin^2(\frac{h}{2})}{8h^2}$.$L = \lim _{h}$ ${ightarrow 0} \frac{1}{2} \cdot \frac{	an(\frac{h}{2})}{\frac{h}{2}} \cdot \frac{2}{8} \cdot (\frac{\sin(\frac{h}{2})}{\frac{h}{2}})^2 \cdot \frac{1}{4}$.$L = \frac{1}{2} \cdot 1 \cdot \frac{1}{4} \cdot 1^2 \cdot \frac{1}{4} = \frac{1}{32}$.

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}} \cdot \frac{1-\sin x}{(\pi-2 x)^3} = $

Similar Questions

The value of $\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{3x - 4}}{{3x + 2}}} \right)^{\frac{{x + 1}}{3}}}$ is equal to

The value of $\lim_{h \rightarrow 0} 2 \left\{ \frac{\sqrt{3} \sin (\frac{\pi}{6} + h) - \cos (\frac{\pi}{6} + h)}{\sqrt{3} h (\sqrt{3} \cos h - \sin h)} \right\}$ is

$\mathop {\lim }\limits_{x \to a} \frac{{{x^2} - {a^2}}}{{x - a}} = $

Find the limit: $\mathop {\lim }\limits_{x \to 1} \frac{x^{2}+1}{x+100}$

Let $f(x) = \lim_{y \rightarrow \infty} y(x^{1/y} - 1)$,and $2022 f(\frac{1}{x}) + P f(x) = f(x^2)$,then $P =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module