Question

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

• A. $\sqrt{2}$
• B. $\frac{\sqrt{2}}{8}$
• C. $0$
• D. none of these

Answer 1

Answer: (B)

$\frac{\sqrt{2}}{8}$

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

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

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

$=\underset{x\to 0}{lim}\frac{2{\sin }^2\frac{x}{2}}{4{\sin }^2\frac{x}{2}.{\cos }^2\frac{x}{2}}.\frac{1}{\sqrt{2}+\sqrt{1+\cos x}}$

$=\underset{x\to 0}{lim}\frac{1}{2{\cos }^2\frac{x}{2}}.\frac{1}{\sqrt{2}+\sqrt{1+\cos x}}$

$=\frac{1}{2}.\frac{1}{\sqrt{2}+\sqrt{2}}=\frac{1}{4\sqrt{2}}=\frac{\sqrt{2}}{8}$