Solution Notice that we have an indeterminate limit case of $\frac{0}{0}, \operatorname{since} \cos 0=1$ and $\sin 0=0.$
We may use the identity
$$
\sin ^{2} x+\cos ^{2} x=1
$$
to obtain
$$
\sin ^{2} x=1-\cos ^{2} x
$$
and write the limit as
$$
\lim _{x \rightarrow 0} \frac{1-\cos x}{\sin ^{2} x}=\lim _{x \rightarrow 0} \frac{1-\cos x}{1-\cos ^{2} x}=\lim _{x \rightarrow 0} \frac{1-\cos x}{(1-\cos x)(1+\cos x)}=\lim _{x \rightarrow 0} \frac{1}{1+\cos x}=\frac{1}{2}.
$$
