Answer
\begin{aligned} \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\sin 2 \theta}= \frac{1}{2}
\end{aligned}
Work Step by Step
Given $$\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\sin 2 \theta} $$
So, we get
\begin{aligned} L&=\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\sin 2 \theta}\\
&= \lim _{\theta \rightarrow 0} \frac{\sin \theta}{1} \frac{1}{\sin 2 \theta} \\
&= \lim _{\theta \rightarrow 0} \frac{\sin \theta}{ \theta} \frac{ \theta}{\sin 2 \theta} \\
&= \lim _{\theta \rightarrow 0} \frac{\sin \theta}{ \theta} \lim _{\theta \rightarrow 0}\frac{ \theta}{\sin 2 \theta} \\
&=1\cdot \lim _{\theta \rightarrow 0}\frac{ \theta}{\sin 2 \theta} \\
&= \frac{1}{2} \lim _{\theta \rightarrow 0}\frac{ 2\theta}{\sin 2 \theta} \\
&= \frac{1}{2}\frac{1}{ \lim \limits _{\theta \rightarrow 0}\frac{ \sin 2\theta}{ 2 \theta} }\\
&= \frac{1}{2}\cdot\frac{1}{ 1 }\\
&= \frac{1}{2}
\end{aligned}
