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