Question

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

• A. $1/2$
• B. $2$
• C. $\sqrt{2}$
• D. Does not exist

Answer 1

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

$\underset{x\to 0}{lim}\frac{\sqrt{2{\sin }^2\frac{x^2}{2}}}{2{\sin }^{2}x/2}$
$=\underset{x\to 0}{lim}\frac{\sqrt{2}\left|\sin \frac{x^2}{2}\right|}{2{\sin }^{2}x/2}$
$=\underset{x\to 0}{lim}\frac{\sin \frac{x^2}{2}}{\sqrt{2}{\sin }^{2}x/2}(\because \sin \frac{x^2}{2}$ is always positive$)$
$=\underset{x\to 0}{lim}\frac{\frac{\sin \frac{x^2}{2}}{\frac{x^2}{2}}}{\frac{\sqrt{2}{\sin }^{2}x/2}{\frac{x^2}{4}}}.\frac{\frac{x^2}{2}}{\frac{x^2}{4}}$
$=\frac{2}{\sqrt{2}}=\sqrt{2}$