_ 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
$\frac{1}{24}$
✓
$\frac{1}{16}$
C
$\frac{1}{8}$
D
$\frac{1}{4}$

✓

Answer

Correct option: B.

$\frac{1}{16}$

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x(1-\sin x)}{(\pi-2 x)^3}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{2}-x\right)\left[1-\cos \left(\frac{\pi}{2}-x\right)\right]}{(\pi-2 x)^3}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{2}-x\right) 2 \sin ^2\left(\frac{\frac{\pi}{2}-x}{2}\right)}{2\left(\frac{\pi}{2}-x\right)(\pi-2 x)^2}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{2}-x\right)}{2\left(\frac{\pi}{2}-x\right)} \cdot \frac{2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{16\left(\frac{\pi}{4}-\frac{x}{2}\right)^2}$
$=\frac{2}{2 \times 16}$
${\left[\because \lim _{x \rightarrow a} \frac{\sin (x-a)}{x-a}=1 \text { and }=\lim _{x \rightarrow a} \frac{\tan (x-a)}{x-a}=1\right]}$
$=\frac{1}{16}$

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