## Calculus (3rd Edition)

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.