Evaluate $\lim_{h \rightarrow 0} {(x+h)^{99}-x^{99}\over h}$ 
$$\lim_{h \rightarrow 0} {(x+h)^{99}-x^{99}\over h}$$

I need to factor this in order to get a limit.
I tried:
$$\lim_{h \rightarrow 0} {[(x+h)^{33}]^3-[(x)^{33}]^3\over h}
\\ \lim_{h \rightarrow 0} {[(x+h)^{33}-(x)^{33}][(x+h)^{66}+(x+h)^{33}(x)^{33}+(x)^{66}]\over h}$$
However this does not seem helpful.
How do I approach this question?

Because both $\lim\limits_{h\to 0}((x-h)^{99}-x^{99})=0$ and $\lim\limits_{h\to 0}h=0$ you may also use L'Hospital's rule:
$\displaystyle\lim\limits_{h\to 0}\frac{(x-h)^{99}-x^{99}}{h}=\displaystyle\lim\limits_{h\to 0}\frac{\frac{d}{dh}((x-h)^{99}-x^{99})}{\frac{d}{dh}h}=\displaystyle\lim\limits_{h\to 0}\frac{99(x-h)^{98}-0}{1}=99x^{98}$
