I am trying to find lim
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!
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