Calculate $\lim_{h\to 0} \frac{\cos(x-2h)-\cos(x+h)}{\sin(x+3h)-\sin(x-h)}$ Calculate $$\lim_{h\to 0} \frac{\cos(x-2h)-\cos(x+h)}{\sin(x+3h)-\sin(x-h)}$$
If I take the limit it results in undefined value.
I try to change the formation using identity $\sin A + \sin B$
$$\lim_{h\to 0} \frac{-2\sin\frac{(2x-h)}2\sin(-3h/2)}{2\cos(x+h)\sin(2h)}$$
How do I actually evaluate the limit? With and without derivative?

Direct evaluation gives $0/0$ so apply L'Hospital's rule:
\begin{align}\lim_{h\to 0}\frac{\frac{d}{dh}\left(\cos(x-2h)-\cos(x+h)\right)}{\frac{d}{dh}\left(\sin(x+3h)-\sin(x-h)\right)}&=\lim_{h\to 0}\frac{2\sin(x-2h)+\sin(x+h)}{3\cos(x+3h)+\cos(x-h)}\\&=\frac{2\sin(x)+\sin(x)}{3\cos(x)+\cos(x)}\\&=\frac{3}{4}\tan(x)
\end{align}
