Question

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

Answer 1

We have,

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

${lim}_{x\to 0}\frac{\sqrt{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}-\sqrt{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}}}{x}$

$={lim}_{x\to 0}\frac{\sqrt{{(\sin \frac{x}{2}+\cos \frac{x}{2})}^2}-\sqrt{{(\sin \frac{x}{2}-\cos \frac{x}{2})}^2}}{x}$

$={lim}_{x\to 0}\frac{\sin \frac{x}{2}+\cos \frac{x}{2}-\sin \frac{x}{2}+\cos \frac{x}{2}}{x}$

$={lim}_{x\to 0}\frac{2\cos \frac{x}{2}}{x}$

$=2{lim}_{x\to 0}\frac{\cos \frac{x}{2}}{\frac{2x}{2}}$

$=\frac{2}{2}{lim}_{x\to 0}\frac{\cos \frac{x}{2}}{\frac{x}{2}}\therefore {lim}_{x\to 0}\frac{\mathrm{cosx}}{x}=1$

$=1\times 1=1$

Hence, this is the answer.