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?

This limit is the derivative of $f(x)=x^{99}$. To get a generalization you can replace $99$ with $n$ where $n \in \mathbb{R}$. The limit thus translates to $\displaystyle \lim_{h \to 0} \frac{(x+h)^{n}-x^n}{h}$. 
Using Binomial Theorem, which states that $(a+b)^n=\displaystyle \sum_{k=0}^n {n \choose k} a^{n-k}\ b^k$. Substituting $\begin{bmatrix}a \\ b\end{bmatrix}=\begin{bmatrix}x \\ h\end{bmatrix}$. $$(x+h)^n=x^n+{n \choose 1}x^{n-1} \ h + {n \choose 2} x^{n-2} h^2+ \cdots +{n \choose n-1}xh^{n-1}+h^n$$ Putting back this result in the expression for the limit. We get the following limit $$\displaystyle \lim_{h \to 0} \frac{(x+h)^n-x^n}{h}=\displaystyle \lim_{h \to 0} \Biggl(\frac{1}{h}{n \choose 1}x^{n-1}h+\frac{1}{h}\biggl(h^2\biggl({n \choose 2}x^{n-2}+{n \choose 3}x^{n-3}h+\cdots+h^{n-2}\biggr)\biggr)\Biggr)$$ $$=\displaystyle \lim_{h \to 0} \Biggl(\biggl(nx^{n-1}\biggr)+h\biggl({n \choose 2}x^{n-2}+{n \choose 3}x^{n-3}h+\cdots+h^{n-2}\biggr)\Biggr)$$ $$=\displaystyle \lim_{h \to 0}nx^{n-1}+\underbrace{\displaystyle \lim_{h \to 0}h\biggl({n \choose 2}x^{n-2}+{n \choose 3}x^{n-3}h+\cdots+h^{n-2}\biggr)}_{=\ 0}$$
$$\implies \displaystyle \lim_{h \to 0}\frac{(x+h)^n-x^n}{h}= nx^{n-1}$$
Therefore, For $n=99$, the given limit evaluates to $99x^{98}$. Cheers
