Question

$\underset{x\to 0}{\lim }\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\left(\sqrt{1+\sin x}-\sqrt{1-\sin x}\right)\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)}{x\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\left(1+\sin x\right)-\left(1-\sin x\right)}{x\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{2\sin x}{x\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)}\right] \\ =\frac{2}{\sqrt{1+0}+\sqrt{1-0}} \\ =\frac{2}{2} \\ =1 \end{gathered}$$