Answer
$$2$$
Work Step by Step
We have
$$
\lim _{x \rightarrow 0} \frac{x^2}{1-\cos x}=\frac{0}{0}
$$
is an intermediate form, then we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 0} \frac{x^2}{1-\cos x}=\lim _{x \rightarrow 0} \frac{2x}{\sin x}=\frac{0}{0}.
$$ Applying L’Hôpital’s Rule, we get $$
\lim _{x \rightarrow 0} \frac{2x}{\sin x}=\lim _{x \rightarrow 0} \frac{2 }{\cos x}=2.
$$
