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

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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} = $

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