_ x 2 x- x ( -2 x)^3 equals

MCQ

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$ equals

A
$\frac{1}{4}$
B
$\frac{1}{24}$
✓
$\frac{1}{16}$
D
$\frac{1}{8}$

✓

Answer

Correct option: C.

$\frac{1}{16}$

(C)
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$
Put $\pi-2 x=\theta$
$\Rightarrow x=\frac{\pi}{2}-\frac{\theta}{2}$ and as $x \rightarrow \frac{\pi}{2}, \theta \rightarrow 0$
$\lim _{\theta \rightarrow 0} \frac{\cot \left(\frac{\pi}{2}-\frac{\theta}{2}\right)-\cos \left(\frac{\pi}{2}-\frac{\theta}{2}\right)}{\theta^3}$
$=\lim _{\theta \rightarrow 0} \frac{\tan \frac{\theta}{2}-\sin \frac{\theta}{2}}{\theta^3}$
$=\lim _{\theta \rightarrow 0} \frac{\sin \frac{\theta}{2}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}}{\cos \frac{\theta}{2} \cdot \theta^3}$
$=\lim _{\theta \rightarrow 0} \frac{\sin \frac{\theta}{2}\left(1-\cos \frac{\theta}{2}\right)}{\cos \frac{\theta}{2} \cdot \theta^3}$
$=\lim _{\theta \rightarrow 0} \frac{\sin \frac{\theta}{2} \cdot 2 \sin ^2 \frac{\theta}{4}}{\cos \frac{\theta}{2} \cdot \theta^3}$
$=2 \lim _{\theta \rightarrow 0} \frac{\sin \frac{\theta}{2}}{\frac{\theta}{2} \times 2} \cdot \frac{\sin ^2 \frac{\theta}{4}}{\left(\frac{\theta}{4}\right)^2 \times 16} \cdot \frac{1}{\cos \frac{\theta}{2}}$
$=\frac{2}{32}=\frac{1}{16}$

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