Answer
$$\frac{4}{3} $$
Work Step by Step
We have
$$
\lim _{x \rightarrow 0} \frac{ \sin4x }{\sin3x}=\lim _{x \rightarrow 0} \frac{4}{3}\frac{ \sin4x }{4x} \frac{ 3x }{\sin3x}\\
= \frac{4}{3}\lim _{4x \rightarrow 0}\frac{ \sin4x }{4x} \lim _{3x \rightarrow 0}\frac{ 3x }{\sin3x}= \frac{4}{3}.
$$
Where we used, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1. $
