Q93 / 200 · Manipal (MET) 2014 · Mathematics
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\left(1-\tan \frac{x}{2}\right)(1-\sin x)}{\left(1+\tan \frac{x}{2}\right)(\pi-2 x)^3}$
is equal to
- A$\frac{1}{8}$
- B0
- Correct answerC$\frac{1}{32}$
- D$\infty$
Solution
Put $x=\frac{\pi}{2}-h$ as $x \rightarrow \frac{\pi}{2}, h \rightarrow 0$
$\begin{aligned}
& \therefore \text { Given, } \operatorname{limit}=\lim _{h \rightarrow 0} \frac{1-\tan \left(\frac{\pi}{4}+\frac{h}{2}\right)(1-\cos h)}{1+\tan \left(\frac{\pi}{4}+\frac{h}{2}\right)(2 h)^3} \\
& \quad=\lim _{h \rightarrow 0} \tan \left(\frac{h}{2}\right) \lim _{h \rightarrow 0} \frac{2 \cdot \sin ^2\left(\frac{h}{2}\right)}{8 h^3} \\
& \quad=\lim _{h \rightarrow 0} \frac{1}{4}\left(\frac{\tan \frac{h}{2}}{2 \times \frac{h}{2}}\right) \times \lim _{h \rightarrow 0}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^2 \times \frac{1}{4} \\
& \\
& \quad=\frac{1}{32}
\end{aligned}$
Hence, the correct answer is option: 'C'.