Answer
$$\lim _{t \rightarrow 0} \frac{\tan 6 t}{\sin 2 t}=3$$
Work Step by Step
Given $$\lim _{t \rightarrow 0} \frac{\tan 6 t}{\sin 2 t}$$
So
\begin{aligned}
\lim _{t \rightarrow 0} \frac{\tan 6 t}{\sin 2 t} &=\lim _{t \rightarrow 0}\left(\frac{\sin 6 t}{t} \cdot \frac{1}{\cos 6 t} \cdot \frac{t}{\sin 2 t}\right)\\
&=\lim _{t \rightarrow 0} \frac{6 \sin 6 t}{6 t} \cdot \lim _{t \rightarrow 0} \frac{1}{\cos 6 t} \cdot \lim _{t \rightarrow 0} \frac{2 t}{2 \sin 2 t} \\ &=6 \lim _{t \rightarrow 0} \frac{\sin 6 t}{6 t} \cdot \lim _{t \rightarrow 0} \frac{1}{\cos 6 t} \cdot \frac{1}{2} \lim _{t \rightarrow 0} \frac{2 t}{\sin 2 t}\\
&=6(1) \cdot \frac{1}{1} \cdot \frac{1}{2}(1)=3
\end{aligned}
