Question

${lim}_{x\to \frac{\pi }{2}}\frac{2^{-\cos x}-1}{x(x-\frac{\pi }{2})}$

Answer 1

We have,

${lim}_{x\to \frac{\pi }{2}}\left(\frac{2^{-\cos x}-1}{x(x-\frac{\pi }{2})}\right)$

This is the $\frac{0}{0}$ form.

So, apply L-Hospital rule

${lim}_{x\to \frac{\pi }{2}}\left(\frac{2^{-\cos x}{\log }_{e}2\times -(-\sin x)-0}{(x-\frac{\pi }{2})+x(1-0)}\right)$

${lim}_{x\to \frac{\pi }{2}}\left(\frac{\sin x\left(2^{-\cos x}{\log }_{e}2\right)}{(x-\frac{\pi }{2})+x}\right)$

$=\frac{\sin \frac{\pi }{2}\left(2^{-\cos \frac{\pi }{2}}{\log }_{e}2\right)}{(\frac{\pi }{2}-\frac{\pi }{2})+\frac{\pi }{2}}$

$=\frac{1\times \left(2^0{\log }_{e}2\right)}{0+\frac{\pi }{2}}$

$=\frac{{\log }_{e}2}{\frac{\pi }{2}}$

$=\frac{2{\log }_{e}2}{\pi }$

$=\frac{{\log }_{e}4}{\pi }$

Hence, this is the answer.