Question

Evaluate: $I=\underset{x\to \frac{\pi }{2}}{\mathrm{lt}}\frac{1-\sin x}{{\left(\pi -2x\right)}^2}$

Answer 1

$\begin{array}{c}{lt}_{x\to \frac{\pi }{2}}\frac{1-\sin x}{{\left(\pi -2x\right)}^2}\\ ={lt}_{h\to 0}\frac{1-\cosh }{{\left(-2h\right)}^2}\left[letx=\frac{\pi }{2}+h\right]\\ ={lt}_{h\to 0}\frac{1-\cosh }{4h^2}=\frac{1}{8}\end{array}$