calculus - Find $lim_{xto0}frac{sin5x}{sin4x}$ using $lim_{thetato0}frac{sintheta}{theta}=1$.

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!