Answer
The limit $\lim _{x \rightarrow 0} \frac{\sin x}{x^{2}}$ does not exist.
Work Step by Step
We have $$
\lim _{x \rightarrow 0} \frac{\sin x}{x^{2}}=\lim _{x \rightarrow 0} \frac{\sin x}{x} \lim _{x \rightarrow 0} \frac{1}{x}= \lim _{x \rightarrow 0} \frac{1}{x}.
$$
Where we used the fact that $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$.
Now, the limit $ \lim _{x \rightarrow 0} \frac{1}{x}$ does not exist. This is because the left and right hand limits are not the same:
$$ \lim _{x \rightarrow 0^+} \frac{1}{x}=\infty,\quad \lim _{x \rightarrow 0^-} \frac{1}{x}=-\infty.$$
Hence, the limit $\lim _{x \rightarrow 0} \frac{\sin x}{x^{2}}$ does not exist.
