This problem is in the context of the section on the mean value theorem: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 $\displaystyle x_0$.

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.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

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

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!