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?
 A: If you're familiar with derivatives, then you can recall that
$$ f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} $$
and consider $f(x) = x^{99}$.
Otherwise, use the binomial theorem:
\begin{align}\frac{(x+h)^{99}-x^{99}}{h} &= \frac{1}{h}\left( \sum_{k=0}^{99} \binom{99}{k}x^{99-k}h^{k} - x^{99} \right) = \frac{1}{h}\sum_{k=1}^{99}\binom{99}{k}x^{99-k}h^{k}\\&= \binom{99}{1}x^{98} + \sum_{k=2}^{99}\binom{99}{k}x^{99-k}h^{k-1}
\to 99x^{98}, \quad \text{as }h\to 0.
\end{align}
A: Hint: That’s basically the derivative of $f(x) = x^{99}$.
$$\lim_{h \rightarrow 0} {(x+h)^{99}-x^{99}\over h}$$
Recall that by Binomial Expansion,
$$(a+b)^n = \sum_{r = 0}^{n} {n \choose r}a^{n-r}b^r$$
So you get
$$\lim_{h \rightarrow 0} {\color{blue}{{99 \choose 0}x^{99}}+{99 \choose 1}x^{98}h+{99 \choose 2}x^{97}h^2+…+{99 \choose n}h^{99}\color{red}{-x^{99}}\over h}$$
Cancel out the first and last terms. From here, notice if anything can be factored. Also, as another hint (for later simplifications), $${n \choose r} = \frac{n!}{r!{(n-r)!}}$$
A: The trick is to realise that $h \to 0$ means $h$ is arbitrarily small so we can assume $h < 1$ and so $h > h^2 > h^3$ and that the "higher values of $h$" become "negligible"
So you are thinking way too hard.  Just use the binomial theorem to get $(x + h)^{99} = x^{99} + 99hx^{98} + \sum_{k=2}^{99} {99 \choose k} h^kx^{99-k}$ and hope that must of those higher values will become negligible.
And indeed:
$\require{cancel}\lim\limits_{h\to 0}\frac {(x + h)^{99}-x^{99}}{h }=$
$\lim\limits_{h\to 0}\frac {x^{99} + 99hx^{98} + (\sum_{k=2}^{99} {99 \choose k} h^kx^{99-k})-x^{99}}{h}=$
$\lim\limits_{h\to 0}\frac {\color{green}{\cancel{x^{99}}} + 99\color{red}{\cancel h}x^{98} + (\sum_{k=2}^{99} {99 \choose k} h^{\color{red}{\cancel {k}}k-1}x^{99-k}) - \color{green}{\cancel{x^{99}}}}{\color{red}{\cancel h}}=$
$\lim\limits_{h\to 0}(99x^{98} + \sum_{k=2}^{99} {99 \choose k}h^{k-1}x^{99-k})=$
$99x^{98} + \sum_{k=2}^{99}{99 \choose k}(\lim\limits_{h\to 0} h^{k-1})x^{99-k}=$
$99x^{98} + \sum_{k=2}^{99}{99 \choose k}(0)x^{99-k}=$
$99x^{98} + \sum_{k=2}^{99}0=$
$99x^{98}$
======
Doug M had an intriguing hint in the comments:
Considering $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b+  .... + ab^{n-2}+ b^{99})$ we get:
$(x + h)^{99} - x^{99}=$
$[(x+h) - x][(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]=$
$h[(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]$
And 
$[(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]=$
$(x^{98} + \text{a bunch of stuff with h as a factor})+(x^{97}\cdot x + x(\text{a bunch of stuff with h as a factor}))+... (x\cdot x^{97} + x^{97}(\text{a bunch of stuff with h as a factor})) + x^{98})=$
$99x^{98} + \text { a REALLY big bunch of stuff with h as a factor}$
So $\frac {(x+h)^{98} - x^{99}}h=$
$\frac {h(99x^{98}+ \text { a REALLY big bunch of stuff with h as a factor}}h = $
$99x^{98}  + \text { a REALLY big bunch of stuff with h as a factor}$
... and "$ \text { a REALLY big bunch of stuff with h as a factor}$" goes to $0$ as .... it's a REALLY big bunch of stuff with $h$ as a factor.
Okay, .... it was an intriguing idea and I'm glad I did it but.... I don't think it's a very practical way to do it.
A: 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:
$\displaystyle\lim\limits_{h\to 0}\frac{(x-h)^{99}-x^{99}}{h}=\displaystyle\lim\limits_{h\to 0}\frac{\frac{d}{dh}((x-h)^{99}-x^{99})}{\frac{d}{dh}h}=\displaystyle\lim\limits_{h\to 0}\frac{99(x-h)^{98}-0}{1}=99x^{98}$
A: 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
