How do I evaluate a statement like $\lim_{x\to a}\Big(\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\Big)$? I need to know how to do this to find out the derivative of a function as close as possible to an undefined point.
Assume that $f$ is not differentiable at $x=a$.
This might be the same as asking how to evaluate 
$\Big(\lim_{h\to 0} \frac{f(x+\delta+h)-f(x+\delta)}{h}\Big) = f'(x+\delta)$
Not sure.
Basically, what's the derivative of a function $f$ at the point closest to $x=a$ if $f$ is not defined at $a$?  Note I'm not asking for the derivative of $f(a)$, since that is undefined, but instead the point closest to the undefined point, the endpoint I suppose.  
What's the endpoint and what's the derivative of it?
I'm guessing I have to break this into a piecewise function but I'd rather not.
 A: Although you already have a couple of answers to your question I would like to add something which you might find clarifying. As other users said, in your case, you should first evaluate the inner limit (that is, if possible, obtain a general expression for $f'(x)$ in a neighbourhood of $a$), and then take the limit for $x \rightarrow a$.  
What I wanted to point out is that, even if you assume that $f$ is differentiable at $x=a$ this is not enough to conclude that  $(1)$ $\lim_{x\to a}\Big(\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\Big)= \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = f'(a)$. Take for example $f(x) = \begin{cases} x^2\sin(\frac{1}{x}) \ \text{if} \ x \neq 0 \newline 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if}\ x=0\end{cases}$ 
You can check that $f$ is differentiable at $x=0$ (through the definition of derivative) but if you compute the derivative $f'(x)$ for $x \neq 0$ and take the limit $\lim_{x \rightarrow 0 }f'(x)$ the you will notice that such limit doesn't exist. In other words $f$ is differentiable at $x = 0$ (and in all of $\mathbb{R}$), but $f'$ isn't continuous at $x=0$. However if the limit $\lim_{x \rightarrow 0 }f'(x)$ did exist, you would be able to conclude that $f$ is differentiable at $x=0$, in other words it is possible to prove the following 
Theorem if $f: \mathbb{R} \rightarrow \mathbb{R}$ is differentiable in a neighbourhood of $a \in \mathbb{R}$, and $\lim_{x \rightarrow a }f'(x)$ exists and is finite, then $f$ is differentiable at $x=a$ and $\lim_{x \rightarrow a }f'(x)= f'(a)$. (Analogous versions hold if the limit isn't finite) 
Concluding $(1)$ without any particular reason (such as the hypothesis of the theorem) is very tempting, but it is also a common mistake in calculus. Hope this helps.
A: First evaluate the limit inside, that is, $\lim_{h\rightarrow 0}\frac{f(x + h) - f(x)}{h}$, and then the limit outside. If $f$ is differentiable at $x$, then the limit inside is just $f'(x)$. So then you're taking the limit $\lim_{x\rightarrow a}f'(x)$. For example, suppose $f(x) = 2x^2 - 1$. The derivative of $f$ is $f'(x) = 4x$, and $\lim_{x\rightarrow a}f'(x) = f'(a) = 4a$.
A: Start with an example.  How about $f(x) = \frac{1}{x}$, which is defined for all $x \neq 0$.  Then, the inner limit is $f'(x) = -\frac{1}{x^2}$, which is also defined for $x \neq 0$.  Taking the limit, $\lim_{x \to 0} -\frac{1}{x^2}$, we get $-\infty$.  You could now start with $g(x) = -\frac{1}{x}$ to find a final limit of $+\infty$.  You could now start with $h(x) = \frac{1}{x^2}$ to end up with left hand and right hand limits that are different, so the limit does not exist at all, as a real number or even an extended real number.
