Question

Find the following limit:
$\underset{x\to 0}{lim}\frac{\sqrt{1-\cos x^2}}{1-\cos x}.$

Answer 1

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

$=\underset{x\to 0}{lim}\frac{\sqrt{2{\sin }^2\frac{x^2}{2}}}{2{\sin }^2\frac{x}{2}}\left[\underset{x\to 0}{lim}\frac{\sin x}{x}=1\right]$

$=\underset{x\to 0}{lim}\frac{\sqrt{2}\cdot \sin \frac{x^2}{2}}{2{\sin }^2\frac{x}{2}}$

$=\underset{x\to 0}{lim}\frac{1}{\sqrt{2}}\frac{\sin \frac{x^2}{2}}{\left(\frac{x^2}{2}\right)}\times \left(\frac{x^2}{2}\right)\times \frac{1}{{\sin }^2\frac{x}{2}}\times \frac{\frac{x^2}{4}}{\frac{x^2}{4}}$

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

$=\frac{1}{\sqrt{2}}\times \frac{x^2}{2}\times \frac{4^2}{x^2}=\frac{2}{\sqrt{2}}=\sqrt{2}$.