Answer
\begin{aligned} \lim _{x \rightarrow 0}\frac{\tan 3 x }{\sin 8 x}= \frac{3}{8}
\end{aligned}
Work Step by Step
Given $$\lim _{x \rightarrow 0} \frac{\tan 3 x }{\sin 8 x} $$
So, we get
\begin{aligned} L&=\lim _{x \rightarrow 0}\frac{\tan 3 x }{\sin 8 x}\\
&= \lim _{x \rightarrow 0} \frac{\tan 3 x}{1} \frac{1}{\sin 8 x} \\
&= \lim _{x \rightarrow 0} \frac{\tan 3 x}{ x} \frac{ x}{\sin 8 x} \\
&= 3\lim _{x\rightarrow 0} \frac{\tan 3 x}{3 x} \cdot \frac{1}{8}\lim _{x \rightarrow 0}\frac{ 8x}{\sin 8 x} \\
&=3\cdot 1 \cdot \frac{1}{8} \lim _{x \rightarrow 0}\frac{ 8x}{\sin 8 x} \\
&= \frac{3}{8}\frac{1}{ \lim \limits _{x \rightarrow 0}\frac{ \sin 8x}{ 8 x} }\\
&= \frac{3}{8}\cdot\frac{1}{ 1 }\\
&= \frac{3}{8}
\end{aligned}
