Finding the derivatives of sin(x) and cos(x) We all know that the following (hopefully):
$$\sin(x)=\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+1}}{(2n+1)!}\ , \ x\in \mathbb{R}$$
$$\cos(x)=\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n}}{(2n)!}\ , \ x\in \mathbb{R}$$
But how do we find the derivates of $\sin(x)$ and $\cos(x)$ by using the definition of a derivative and the those definition above?
Like I should start by doing:
$$\lim_{h\to\infty}\frac{\sin(x+h)-\sin(x)}{h}$$
But after that no clue at all.  Help appreciated!

The radius of convergence of the power series is infinity, so you can interchange summation and differentiation.
$\frac{d}{dx}\sin x = \frac{d}{dx} \sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+1}}{(2n+1)!} = \sum^{\infty}_{n=0}(-1)^n \frac{d}{dx} \frac{x^{2n+1}}{(2n+1)!} = \sum^{\infty}_{n=0}(-1)^n  \frac{x^{2n}}{(2n)!} = \cos x$
If you wish to use the definition, you can do:
$\frac{\sin(x+h)-\sin x}{h} = \frac{\sum^{\infty}_{n=0}(-1)^n \frac{(x+h)^{2n+1}}{(2n+1)!}-\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+1}}{(2n+1)!}}{h} = \sum^{\infty}_{n=0}(-1)^n \frac{1}{(2n+1)!}\frac{(x+h)^{(2n+1)}-x^{(2n+1)}}{h}$
Then $\lim_{h \to 0} \frac{(x+h)^{(2n+1)}-x^{(2n+1)}}{h} = (2n+1)x^{2n}$ (using the binomial theorem).
(Note $(x+h)^p-x^p = \sum_{k=1}^p \binom{p}{k}x^{p-k}h^k = p x^{p-1}h + \sum_{k=2}^p \binom{p}{k}x^{p-k}h^k$. Dividing across by $h$ and taking limits as $h \to 0$ shows that $\frac{d}{dx} x^p = p x^{p-1}$.)
