Question

$\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{\sqrt{2}-\sqrt{1+\sin x}}{{\cos }^{2}x}$

Answer 1

$$\begin{gathered} \underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\left[\frac{\sqrt{2}-\sqrt{1+\sin x}}{{\cos }^{2}x}\right] \\ \mathrm{Rationalising}\mathrm{the}\mathrm{numerator}: \\ =\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\left[\frac{\left(\sqrt{2}-\sqrt{1+\sin x}\right)\left(\sqrt{2}+\sqrt{1+\sin x}\right)}{{\cos }^{2}x\times \left(\sqrt{2}+\sqrt{1+\sin x}\right)}\right] \\ =\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\left[\frac{2-\left(1+\sin x\right)}{{\cos }^{2}x·\left(\sqrt{2}+\sqrt{1+\sin x}\right)}\right] \\ =\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\left[\frac{\left(1-\sin x\right)}{\left(1-\sin x\right)\left(1+\sin x\right)\left(\sqrt{2}+\sqrt{1+\sin x}\right)}\right] \\ =\frac{1}{\left(1+1\right)\left(\sqrt{2}+\sqrt{2}\right)} \\ =\frac{1}{2\times 2\sqrt{2}} \\ =\frac{1}{4\sqrt{2}} \end{gathered}$$