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

\(\begin{aligned} \text { Let } \mathrm{I} & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3} \\ & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x(1-\sin x)}{\sin x(\pi-2 x)^3} \\ \text { Put } x & =\frac{\pi}{2}-\mathrm{h} \\ \therefore \quad \pi-2 x & =2 \mathrm{~h} \\ \text { As } x & \rightarrow \frac{\pi}{2}, \mathrm{~h} \rightarrow 0\end{aligned}\)
\(\begin{aligned} \therefore \quad I & =\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{2}-h\right)\left(1-\sin \left(\frac{\pi}{2}-h\right)\right)}{\sin \left(\frac{\pi}{2}-h\right)(2 h)^3} \\ & =\lim _{h \rightarrow 0} \frac{\sin h \cdot 2 \sin ^2\left(\frac{h}{2}\right)}{\cosh \cdot 8 h^3} \\ & =\frac{2}{8} \lim _{h \rightarrow 0} \frac{\sin h}{h} \times \lim _{h \rightarrow 0} \frac{\sin ^2 \frac{h}{2}}{\frac{h^2}{4} \cdot 4} \\ & =\frac{2}{8}(1) \times(1) \times \frac{1}{4} \\ & =\frac{1}{16}\end{aligned}\)