$\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}-\sqrt{1+\cos x}}$ equals :
Correct Option: 1
$\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}-\sqrt{1+\cos x}}$
$=\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}-\sqrt{2 \cos ^{2} \frac{x}{2}}} \quad\left[\frac{0}{0}\right]$ $\left[\frac{0}{0}\right]$
$=\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}\left[1-\cos \frac{x}{2}\right]}=\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{2 \sqrt{2} \sin ^{2} \frac{x}{4}}$
$=\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}\left[1-\cos \frac{x}{2}\right]}=\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{2 \sqrt{2} \sin ^{2} \frac{x}{4}}$
$=\lim _{x \rightarrow 0} \frac{\left(\frac{\sin x}{x}\right)^{2} \cdot 16}{2 \sqrt{2}\left(\frac{\sin \frac{x}{4}}{\frac{x}{4}}\right)^{2}}=\frac{16}{2 \sqrt{2}}=4 \sqrt{2}$