Question

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

Answer 1

$$\begin{gathered} \underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{1-\sin x}{{\left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}^2} \\ =\underset{h\to 0}{\lim }\frac{1-\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)}{{\left({\displaystyle \frac{\mathrm{\pi }}{2}}-\left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)\right)}^2} \\ =\underset{h\to 0}{\lim }\frac{1-\cos h}{h^2} \\ =\underset{h\to 0}{\lim }\frac{2{\sin }^2{\displaystyle \frac{h}{2}}}{{\displaystyle \frac{4h^2}{4}}}\left[\because \underset{h\to 0}{\lim }\frac{\sin h}{h}=1\right] \\ =\frac{1}{2}\underset{h\to 0}{\lim }{\left(\frac{\sin {\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h}{2}}}\right)}^2 \\ \Rightarrow \frac{1}{2} \end{gathered}$$