Question

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

Answer 1

$$\begin{gathered} \underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{\sqrt{2-\sin x}-1}{{\left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}^2} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{2-\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)}-1}{{\left({\displaystyle \frac{\mathrm{\pi }}{2}}-\left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)\right)}^2} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{2-\cos h}-1}{h^2} \\ \text{Dividing the numerator and the}\mathrm{denominator}\mathrm{by}\sqrt{2-\cos h}+1: \\ \underset{h\to 0}{\lim }\frac{\left(\sqrt{2-\cos h}-1\right)\left(\sqrt{2-\cos h}+1\right)}{\left(\sqrt{2-\cos h}+1\right)h^2} \\ =\underset{h\to 0}{\lim }\frac{2-\cos h-1}{h^2\left(\sqrt{2-\cos h}+1\right)} \\ =\underset{h\to 0}{\lim }\frac{1-\cos h}{h^2\left[\sqrt{2-\cos h}+1\right]} \\ =\underset{h\to 0}{\lim }\frac{2{\sin }^2{\displaystyle \frac{h}{2}}}{{\displaystyle \frac{4h^2}{4}}\left(\sqrt{2-\cos h}+1\right)} \\ =\frac{1}{2}\underset{h\to 0}{\lim }{\left(\frac{\sin {\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h}{2}}}\right)}^2\times \underset{h\to 0}{\lim }\left(\frac{1}{\sqrt{2-\cos h}+1}\right) \\ =\frac{1}{2\left(\sqrt{2-1}+1\right)} \\ =\frac{1}{4} \end{gathered}$$