Answer
The limit does not exist.
Work Step by Step
By substitution, we get the intermediate form $\frac{0}{0}$, so we have
$$
\lim _{t \rightarrow 0} \frac{\sin^2 t}{t^2 } \frac{1}{t}=\lim _{t \rightarrow 0} \frac{\sin^2 t}{t^2 } \lim _{t \rightarrow 0} \frac{1}{t}\\
=(1)\frac{1}{0}.
$$
We check the one-sided limits
$$
\lim _{t \rightarrow 0^-} \frac{1}{t } =\frac{1}{0^-}=-\infty
$$
$$
\lim _{t \rightarrow 0^+} \frac{1}{t } =\frac{1}{0^+}=\infty
$$
so the limit does not exist.
