## Calculus (3rd Edition)

$$\frac{4}{3}$$
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, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1.$