Find $\lim_{x\to0}\frac{\sin5x}{\sin4x}$ using $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$. I am trying to find $$\lim_{x\to0}\frac{\sin5x}{\sin4x}$$
My approach is to break up the numerator into $4x+x$. So,
$$\begin{equation*}
\lim_{x\to0}\frac{\sin(4x+x)}{\sin4x}=\lim_{x\to0}\frac{\sin4x\cos x+\cos4x\sin x}{\sin4x}\\
=\lim_{x\to0}(\cos x +\cos4x\cdot\frac{\sin x}{\sin4x})\end{equation*}$$
Now the problem is with $\frac{\sin x}{\sin4x}$. If I use the double angle formula twice, it is going to complicate the problem.
The hint says that you can use $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$.
I have little clue how can I make use of the hint.
Any helps are greatly appreciated. Thanks!

$$\lim_{x\to0}\frac{\sin5x}{\sin4x}=\lim_{x\to0}\frac{5x\frac{\sin5x}{5x}}{4x\frac{\sin4x}{4x}}=\frac{5}{4}\frac{\lim_{x\to0}\frac{\sin5x}{5x}}{\lim_{x\to0}\frac{\sin4x}{4x}}=\frac{5}{4}$$
