Q101 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Mathematics

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

  • Correct answer
    A
    $\frac{1}{32}$
  • B
    $0$
  • C
    $\frac{1}{16}$
  • D
    $\frac{1}{8}$

Solution

$\begin{aligned} & \lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}} \cdot \frac{1-\sin x}{(\pi-2 x)^3} \\ & =\lim _{x \rightarrow \frac{\pi}{2}} \tan \left(\frac{\pi}{4}-\frac{x}{2}\right): \frac{1-\sin x}{(\pi-2 x)^3} \\ & \text { On putting } x=\frac{\pi}{2}-y, y \rightarrow 0 \\ & =\lim _{y \rightarrow 0} \frac{\tan \left(\frac{\pi}{4}-\frac{\frac{\pi}{2}-y}{2}\right) \cdot\left(1-\sin \left(\frac{\pi}{2}-y\right)\right.}{\left(\pi-2\left(\frac{\pi}{2}-y\right)\right)^3} \\ & =\lim _{y \rightarrow 0} \frac{\left(\tan \frac{y}{2}\right) \cdot(1-\cos y)}{8 y^3} \\ & =\lim _{y \rightarrow 0} \frac{\left(\tan \frac{y}{2}\right) \cdot 2 \cdot\left(\sin \frac{y}{2}\right)\left(\sin \frac{y}{2}\right)}{64 \cdot\left(\frac{y}{2}\right)\left(\frac{y}{2}\right)\left(\frac{y}{2}\right)} \\ & =\frac{2}{64}=\frac{1}{32} \\ & \end{aligned}$

Hence, the correct answer is option: 'A'.

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