#### Find the limit of the following:  $\lim _{x \rightarrow \pi / 2}[1+(\cot x-\cos x)]^{1 /(\pi-2 x)^3}$Option: 1 $e^{\frac{1}{16}}$Option: 2 $0$Option: 3 $\frac{2}{n}$Option: 4 $e$

Given: $\lim _{x \rightarrow \pi / 2}[1+(\cot x-\cos x)]^{1 /(\pi-2 x)^3}$

If $\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} g(x)=0, then e^{\lim _{x \rightarrow a} \frac{f(x)}{g(x)}}$.

$=e^{\lim _{x \rightarrow \pi / 2} \frac{\cot x-\cos x}{(\pi-2 x)^3}}$
$=e^{\lim _{x \rightarrow \pi / 2} \frac{\cot x(1-\sin x)}{-8(x-\pi / 2)^3}}$

Apply the trigonometric ratios $\cot x=\tan \left(90^{\sharp}-x\right) and \sin x=\cos \left(90^{\sharp}-x\right)$,the equation becomes,

$=e^{\lim _{x \rightarrow \pi / 2} \frac{\tan (\pi / 2-x)\left[1-\cos \left(\pi / 2^{-x}\right)\right]}{8\left(\pi / 2^{-x}\right)(\pi / 2-x)^2}}$

Again, use the trigonometric formula $1-\cos 2 x=2 \sin ^2 x$, the equation becomes,

$=e^{\lim _{x \rightarrow \pi/ 2} \frac{\tan \left(\pi / 2^{-x}\right)\left[\sin ^2(\pi / 2-x)\right]}{8(\pi / 2-x) 2(\pi / 2-x)^2}}$
$=e^{1 / 16}$

Use the limit formula $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$.

$= e^{\lim_{x\rightarrow \pi }2\cos \left ( x/2 \right )}$
$= e^{0}$
$= 1$

Therefore, $\lim _{x \rightarrow \pi / 2}[1+(\cot x-\cos x)]^{1 /(\pi-2 x)^3}= e^{1/16}.$

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