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 $4\sqrt{2}$

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