Question

$\underset{x\to \mathrm{\pi }/2}{\lim }\frac{2^{-\cos x}-1}{x\left(x-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}$

Answer 1

$$\begin{gathered} \underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\left[\frac{2^{-\cos x}-1}{x\left(x-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}\right] \\ =\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\left[\frac{2^{-\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}-1}{x\left(x-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}\right]\left\{\because \cos x=\sin \left(\frac{\mathrm{\pi }}{2}-x\right)\right\} \\ =\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\left[\frac{2^{\sin \left(x-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}-1}{\left(x-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)\times x}\right] \\ =\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\left[\frac{2^{\sin \left(x-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}-1}{\sin \left(x-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}\times \frac{\sin \left(x-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{\left(x-{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}\times \frac{1}{x}\right] \\ ={\log }_{e}2\times 1\times \frac{1}{{\displaystyle \frac{\mathrm{\pi }}{2}}} \\ =\frac{2}{\mathrm{\pi }}{\log }_{e}2 \end{gathered}$$