Question

The value of $\underset{x\to 0}{lim}\frac{(\sqrt{2+x\sin x}-\sqrt{2\cos x})(1+\cos x)}{1-\cos x}$ is

• A. $2\sqrt{2}$
• B. $\sqrt{2}$
• C. $\frac{1}{2\sqrt{2}}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

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

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

$=\underset{x\to 0}{lim}\frac{\sqrt{2+x\sin x}-\sqrt{2\cos x}}{1-\cos x}\frac{\sqrt{2+x\sin x}+\sqrt{2\cos x}}{\sqrt{2+x\sin x}+\sqrt{2\cos x}}\times \underset{x\to 0}{lim}(1+\cos x)$

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

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

$=\sqrt{2}(1+1)$
$=2\sqrt{2}$