Answer
$$2$$
Work Step by Step
Given $$\lim_{\theta\to 0 }\frac{\theta^2}{1-\cos\theta}$$
Then
\begin{align*}
\lim_{\theta\to 0 }\frac{\theta^2}{1-\cos\theta}&=\lim_{\theta\to 0 }\frac{\theta^2}{2\sin^2(\theta/2)}\\
&=\frac{1}{2}\frac{\lim_{\theta\to 0 }(1)}{ \lim_{\theta\to 0 }\frac{\sin^2(\theta/2)}{\theta^2 } }\\
&=\frac{1}{2}\frac{\lim_{\theta\to 0 }(1)}{ \frac{1}{4} \lim_{\theta\to 0 }\left(\frac{\sin(\theta/2)}{\theta/2 }\right)^2 }\\
&=\frac{1/2}{1/4}\\
&=2
\end{align*}
