Question

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

• A. $4\sqrt{2}$
• B. $4$
• C. $\sqrt{2}$
• D. $2\sqrt{2}$

Answer 1

The correct option is A $4\sqrt{2}$

$L=\underset{x\to 0}{lim}\frac{{\sin }^{2}x}{\sqrt{2}-\sqrt{1+\cos x}}$
$=\underset{x\to 0}{lim}\frac{{\sin }^{2}x}{\sqrt{2}-\sqrt{2{\cos }^2\frac{x}{2}}}$
$=\underset{x\to 0}{lim}\frac{{\sin }^{2}x}{\sqrt{2}[1-\cos \frac{x}{2}]}=\underset{x\to 0}{lim}\frac{{\sin }^{2}x}{2\sqrt{2}{\sin }^2\frac{x}{4}}=\underset{x\to 0}{lim}\frac{1}{2\sqrt{2}}.\left(\frac{{\sin }^{2}x}{x^2}\right).\left(\frac{(x/4{)}^2}{{\sin }^2\frac{x}{4}}\right).16=\frac{16}{2\sqrt{2}}=4\sqrt{2}$