Answer
$$\infty$$
Work Step by Step
Given $$\lim_{\theta\to 0^+}\frac{\sin \theta }{\theta^2}$$
Then
\begin{align*}
\lim_{\theta\to 0^+}\frac{\sin \theta }{\theta^2}&= \lim_{\theta\to 0^+}\frac{\sin \theta }{\theta }\frac{1 }{\theta } \\
&= \lim_{\theta\to 0^+}\frac{\sin \theta }{\theta } \lim_{\theta\to 0^+}\frac{1 }{\theta } \\
&=(1)(\infty)\\
&=\infty
\end{align*}