Calculus: Find the Limit of (f(x+h)-f(x))/h as h approaches 0 for f(x)=cos(2x) Find the limit:
$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
Given that $f(x) = \cos(2x)$
Tried many ways, but I kept on getting an indeterminate form. I can't find a way to cancel out terms on the numerator and denominator.
Any help will be appreciated.

The OP posted a comment below nbubis's answer asking what to do with $\frac{\sin^2(\frac{h}{2})}{h}$. I can't comment back, (no reputation at all) so here's my contribution, multiply $\frac{h}{4}$  in the numerator and denominator. This will lead to $\frac{\sin^2(\frac{h}{2})}{(\frac{h}{2})^2}*\frac{h}{4}$. The first term is just one squared, the second term is $0$. So the term evaluates to zero. Accept his answer.
EDIT: I decided to post my own version of the answer anyway.
We know that, $\cos(C)-\cos(D)=2\sin(\frac{C+D}{2})\sin(\frac{D-C}{2})$
Now, Let $C=2x+h, D=2x$
Hence, our limit becomes, $\frac{2\sin(\frac{4x+h}{2})sin(\frac{-h}{2})}{\frac{h}{2}}$
Now, $\frac{\sin(\frac{-h}{2})}{h}$ evaluates to $-\frac{1}{2}$. Proceed.
