is the Derivative's power rule wrong? I'm studying about the proof of Derivative's power rule and confuse in the algebra of this limit:
consider : $ f(x) = x^n $ , $n = 0, 1, 2, 3, ...$
$n=0 : f(x) = x^0 = 1 $ (where x not equal 0)
$ f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} $
$ = \lim_{h\to 0} \frac{1-1}{h} = 0 $
Why is this limit of $h$ when $h$ approaches $0$ equal to $0$ ????  it has to be undefined as h approach $0$ then the equation become $\frac{0}{0}$
or this proof is wrong?
 A: Let's consider a less abstract example first: let $f(x)=0$ if $x\not=0$, and $1$ if $x=0$. What's the limit as $x$ approaches zero of $f(x)$? The limit of a function at a point depends on the value of the function near, but not at, the point; so this limit is just $0$. The fact that $f(0)$ is not equal to $\lim_{x\rightarrow 0}f(x)$ just means $f$ is not nicely behaved - the technical term here is continuous - at zero.

The example you ask about isn't really any different - instead of being defined differently at $h=0$, it's undefined at $h=0$, but that doesn't change the analysis. ${1-1\over h}=0$ for all $h\not=0$, so the limit as $h$ approaches zero is $0$.
A: When taking a limit we do not care about the value of $x$ is when it reaches $0$, we only care about the values as it gets arbitrarily close to $0$. 
Another way to look at it is that taking a limit is a function so you get something like this: 
$  \lim_{h\to 0} (f(x))= \lim_{h\to 0} (\frac{1-1}{h}) = \lim_{h\to 0} (0) = 0  $
You should also look into limit properties for example, $ \lim_{x\to c} b =b$. Also you said that "$\frac{0}{0}$ has to be undefined". It is not we call such a term indeterminate which is different.
Here is a list of limit properties with exercises incase you want to know more: http://tutorial.math.lamar.edu/Classes/CalcI/LimitsProperties.aspx
Hope this helps!
