Answer
$$ \lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\sin 2 \theta}=0 $$
Work Step by Step
Given $$ \lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\sin 2 \theta} $$
So, we get
\begin{aligned} L&= \lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\sin 2 \theta} \\
&=\lim _{\theta \rightarrow 0}\left(\frac{\theta}{\sin 2 \theta} \cdot \frac{1-\cos \theta}{\theta}\right)\\ &=\lim _{\theta \rightarrow 0} \frac{\theta}{\sin 2 \theta} \cdot \lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta}\\ &=\lim _{\theta \rightarrow 0}\left(\frac{1}{2} \cdot \frac{2 \theta}{\sin 2 \theta}\right) \cdot \lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta}\\ &=\frac{1}{2} \lim _{\theta \rightarrow 0} \frac{2 \theta}{\sin 2 \theta} \cdot \lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta}\\ &=\frac{1}{2} \cdot 1 \cdot 0
\end{aligned}
Where, \begin{aligned} \lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta}&= \lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta}\frac{1+\cos \theta}{1+\cos \theta}\\
&= \lim _{\theta \rightarrow 0} \frac{1-\cos^2 \theta}{\theta(1+\cos \theta)}\\
&= \lim _{\theta \rightarrow 0} \frac{\sin^2 \theta}{\theta(1+\cos \theta)}\\
&= \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} \cdot\lim _{\theta \rightarrow 0} \sin \theta \cdot\lim _{\theta \rightarrow 0} \frac{1}{(1+\cos \theta)}\\
&= 1\cdot\sin 0\cdot \frac{1}{(1+\cos 0)}\\
&=0
\end{aligned}
