Q:

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?

A:

Now, $$\frac{-2\sin\frac{2x-h}2\sin(-\frac{3h}{2})}{2\cos(x+h)\sin2h}= \frac{-2\sin\frac{(2x-h)}2\frac{\sin(-\frac{-3h}{2})}{-\frac{3h}{2}}}{2\cos(x+h)\frac{\sin2h}{2h}}\cdot\frac{-\frac{3}{2}}{2}\rightarrow\frac{3\sin{x}}{4\cos{x}}$$

A:

You need to use $$\lim_{y\to 0}\frac{\sin y}y=1$$
Then $$\lim_{h\to 0}\frac{\sin(-3h/2)}{\sin(2h)}=\lim_{h\to 0}\frac{\sin(-3h/2)}{-3h/2}\frac{2h}{\sin(2h)}\frac{-3h/2}{2h}=-\frac 34$$

