Prove f is differentiable at x0 if limit of f' exists at x0

Quote:

Suppose $\displaystyle f$ is differentiable on $\displaystyle (a,b)$, except possibly at $\displaystyle x_0\in(a,b)$, and is continuous on $\displaystyle [a,b]$; assume $\displaystyle \lim_{x\to x_0}f'(x)$ exists. Prove that $\displaystyle f$ is differentiable at $\displaystyle x_0$ and $\displaystyle f'$ is continuous at $\displaystyle x_0$.

This problem is in the context of the section on the mean value theorem:

Quote:

If $\displaystyle f:[a,b]\to R$ is continuous on $\displaystyle [a,b]$ and differentiable on $\displaystyle (a,b)$, then there is a $\displaystyle c\in (a,b)$ such that

$\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}$

The section also talks about Rolle's Theorem and the Cauchy Mean Value Theorem, but they do not appear to be relevant to this problem.

I've tried attacking this proof from several angles, but I just don't see how the mean value theorem could fit in, or how otherwise to prove it.

Here's what I most recently tried...

Choose $\displaystyle \epsilon>0$ and $\displaystyle \delta=\min\{\delta_i\}$ (where $\displaystyle \delta_i$ will be shortly defined).

Observe that for each $\displaystyle x,\alpha\in(x_0-\delta,x_0)$ and for each $\displaystyle x,\alpha\in(x_0,x_0+\delta)$, there is $\displaystyle c\in(x,\alpha)$ or $\displaystyle c\in(\alpha,x)$ with

$\displaystyle f'(c)=\frac{f(x)-f(\alpha)}{x-\alpha}$ $\displaystyle \Rightarrow$ $\displaystyle f'(c)(x-\alpha)=f(x)-f(\alpha)$

Now, we know that

$\displaystyle \exists\; L=\lim_{x\to x_0}f'(x)$

Which is to say that there is $\displaystyle \delta_2>0$ such that if $\displaystyle 0<|x-x_0|<\delta$ then $\displaystyle |f'(x)-L|<\epsilon$, or

$\displaystyle |\lim_{\alpha\to x}\frac{f(x)-f(\alpha)}{x-\alpha}-L|<\epsilon$ $\displaystyle \Rightarrow$ $\displaystyle |f'(c)-L|<\epsilon$

But that just has me going in circles.

Any help would be much appreciated!