$\lim _{x ightarrow \frac{\pi}{2}} \frac{\left(1- an \left(\frac{x}{2} ight) ight)(1-\sin x)}{\left(1+ an \left(\frac{x}{2} ight) ight)(\pi-2 x)^3}$ is

Question

$\lim _{x ightarrow \frac{\pi}{2}} \frac{\left(1-	an \left(\frac{x}{2}ight)ight)(1-\sin x)}{\left(1+	an \left(\frac{x}{2}ight)ight)(\pi-2 x)^3}$ is

Aicharya · Accepted Answer

$\begin{aligned} 	ext {Let } l & =\lim _{x ightarrow \frac{\pi}{2}}\left[\frac{1-	an \left(\frac{x}{2}ight)}{1+	an \left(\frac{x}{2}ight)}ight]\left[\frac{(1-\sin x)}{(\pi-2 x)^3}ight] \ & =\lim _{x ightarrow \frac{\pi}{2}} \frac{\left.	an \left(\frac{\pi}{4}-\frac{x}{2}ight)ight](1-\sin x)}{(\pi-2 x)^3} \end{aligned}$ Put $\pi-2 x=	heta$ $\Rightarrow x=\frac{\pi}{2}-\frac{	heta}{2}$ and as $x ightarrow \frac{\pi}{2}, 	heta ightarrow 0$ $\begin{aligned} 	herefore \quad l & =\lim _{	heta ightarrow 0} \frac{	an \frac{	heta}{4}\left(1-\cos \frac{	heta}{2}ight)}{	heta^3} \ & =\lim _{	heta ightarrow 0} \frac{	an \frac{	heta}{4}}{\frac{	heta}{4} 	imes 4} \cdot \frac{2 \sin ^2 \frac{	heta}{4}}{\frac{	heta^2}{16} 	imes 16} \ & =\frac{1}{32} \lim _{	heta ightarrow 0}\left[\frac{	an \frac{	heta}{4}}{\frac{	heta}{4}} \cdot\left(\frac{\sin \frac{	heta}{4}}{\frac{	heta}{4}}ight)^2ight]=\frac{1}{32}(1)(1)^2=\frac{1}{32}\end{aligned}$

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

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