Solve the following

Question:

$\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}-\sqrt{1+\cos x}}$ equals :

  1. (1) $4 \sqrt{2}$

  2. (2) $\sqrt{2}$

  3. (3) $2 \sqrt{2}$

  4. (4) 4


Correct Option: 1

Solution:

$\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}-\sqrt{1+\cos x}}$

$=\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}-\sqrt{2 \cos ^{2} \frac{x}{2}}} \quad\left[\frac{0}{0}\right]$ $\left[\frac{0}{0}\right]$

$=\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}\left[1-\cos \frac{x}{2}\right]}=\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{2 \sqrt{2} \sin ^{2} \frac{x}{4}}$

$=\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}\left[1-\cos \frac{x}{2}\right]}=\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{2 \sqrt{2} \sin ^{2} \frac{x}{4}}$

$=\lim _{x \rightarrow 0} \frac{\left(\frac{\sin x}{x}\right)^{2} \cdot 16}{2 \sqrt{2}\left(\frac{\sin \frac{x}{4}}{\frac{x}{4}}\right)^{2}}=\frac{16}{2 \sqrt{2}}=4 \sqrt{2}$

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