Question

Find the following limits
$\underset{x\to 0}{lim}\frac{x^2}{1-cosx}$

Answer 1

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

We see that this limit is of $\frac{0}{0}$ form.

Hence, applying $L$ $Hopital{s}^{\prime }$ rule-

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

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

$=\frac{2}{1}=2$