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!

Actually it won't complicate the problem. You can get the same result by using formula 
$$\sin(2x)= 2\sin(x)\cos(x)$$
i.e. in the denominator of second term expand $\sin(4x)$ as $$2\sin(2x)\cos(2x) = 4\sin(x)\cos(x)\cos(2x).$$ By cancelling sin(x) in numerator and denominator you get a whole function in cosine terms. Just substitute zero and you get the limit.
