Answer
\begin{aligned} \lim _{x \rightarrow 0} \frac{\sin 5 x }{\sin 4 x}=\frac{5}{4}
\end{aligned}
Work Step by Step
Given $$\lim _{x \rightarrow 0} \frac{\sin 5 x }{\sin 4 x} $$
So, we get
\begin{aligned} L&=\lim _{x \rightarrow 0} \frac{\sin 5 x }{\sin 4 x}\\
&= \lim _{x \rightarrow 0} \frac{\sin 5 x}{1} \frac{1}{\sin 4 x} \\
&= \lim _{x \rightarrow 0} \frac{\sin 5 x}{ x} \frac{ x}{\sin 4 x} \\
&= 5\lim _{x\rightarrow 0} \frac{\sin 5x}{5 x} \cdot \frac{1}{4}\lim _{x \rightarrow 0}\frac{ 4x}{\sin 4 x} \\
&=5\cdot 1 \cdot \frac{1}{4} \lim _{x \rightarrow 0}\frac{ 4x}{\sin 4 x} \\
&= \frac{5}{4}\frac{1}{ \lim \limits _{x \rightarrow 0}\frac{ \sin 4x}{ 4 x} }\\
&= \frac{5}{4}\cdot\frac{1}{ 1 }\\
&= \frac{5}{4}
\end{aligned}
