Question

Evaluate the limit:
$\underset{x\to \pi /2}{lim}\frac{2^{-\cos x}-1}{x(x-\frac{\pi }{2})}$

Answer 1

Given:
$\underset{x\to \pi /2}{lim}\frac{2^{-\cos x}-1}{x(x-\frac{\pi }{2})}$

$=\underset{x\to \frac{\pi }{2}}{lim}\left(\frac{2^{-\sin (\frac{\pi }{2}-x)}-1}{x(x-\frac{\pi }{2})}\right)$

$[\because \cos x=\sin (x-\frac{\pi }{2})]$

$=\underset{x\to \frac{\pi }{2}}{lim}\left(\frac{2^{-\sin (\frac{\pi }{2}-x)}-1}{(x-\frac{\pi }{2})\times x}\right)$

$=\underset{x\to \frac{\pi }{2}}{lim}(\frac{2^{-\sin (x-\frac{\pi }{2})}-1}{\sin (x-\frac{\pi }{2})}\times \frac{\sin (x-\frac{\pi }{2})}{(x-\frac{\pi }{2})}\times \frac{1}{x})$

$=\underset{(x-\frac{\pi }{2})\to 0}{lim}\left(\frac{2^{-\sin (x-\frac{\pi }{2})}-1}{\sin (x-\frac{\pi }{2})}\right)\times 1\times \frac{1}{\frac{\pi }{2}}\because [{lim}_{x\to 0}\frac{\sin x}{x}=1]$

$=\log 2\times \frac{2}{\pi }$

$[\because \underset{x\to 0}{lim}\left(\frac{a^x-1}{x}\right)=\log a]$

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