Prove that $f'(a)=\lim_{x\rightarrow a}f'(x)$. Let $f$ be a real-valued function continuous on $[a,b]$ and differentiable on $(a,b)$.
Suppose that $\lim_{x\rightarrow a}f'(x)$ exists.
Then, prove that $f$ is differentiable at $a$ and $f'(a)=\lim_{x\rightarrow a}f'(x)$.  
It seems like an easy example, but a little bit tricky.
I'm not sure which theorems should be used in here.  
==============================================================
Using @David Mitra's advice and @Pete L. Clark's notes
I tried to solve this proof.
I want to know my proof is correct or not.
By MVT, for $h>0$ and $c_h \in (a,a+h)$
$$\frac{f(a+h)-f(a)}{h}=f'(c_h)$$
and $\lim_{h \rightarrow 0^+}c_h=a$.  
Then $$\lim_{h \rightarrow 0^+}\frac{f(a+h)-f(a)}{h}=\lim_{h \rightarrow 0^+}f'(c_h)=\lim_{h \rightarrow 0^+}f'(a)$$ 
But that's enough? I think I should show something more, but don't know what it is.  
 A: The result is essentially Theorem 5.29 from my honors calculus notes.  As I mention, I learned this result from Spivak's Calculus.  I say "essentially" because the version discussed in my notes is for an interior point of an interval whereas your version is at an endpoint, but to prove the two-sided version you just make a one-sided argument twice, so it's really the same thing.
[And David Mitra is right: the proof uses the Mean Value Theorem and not much else.]
Added: Since we are talking about this result anyway: although I call it a "theorem of Spivak", this is not entirely serious -- the result is presumably much older than Michael Spivak.  I am just identifying my (probably secondary) source.  If someone knows a primary source, I'd be very happy to hear it.
Also it is of interest to ask what this result is used for.  In my notes it isn't used for anything but is only a curiosity.  I think Spivak does use it for something, though I forget what at the moment.  Moreover a colleague of mine called my attention to this result in the context of, IIRC, Taylor's Theorem.  Does anyone know of further applications?
A: You need to show that the limit
$$
\lim_{h\to 0^+}
\frac{f(a+h)-f(a)}{h}
$$
exists and it is equal to $\,L=\lim_{x\to a^+}{f'(x)}$.
Let $\varepsilon>0$. We need to find a $\delta>0$, such that
$$
0<h<\delta\quad\Longrightarrow\quad \left|\,\frac{f(a+h)-f(a)}{h}-L\,\right|<\varepsilon.
$$
For this $\,\varepsilon>0,\,$ there exists a $\delta>0$, such that
$$
0<h<\delta\quad\Longrightarrow\quad \left|\,\,f'(a+h)-L\,\right|<\varepsilon.
$$
At the same time, for every $h\in (0,\delta)$, according to the Mean Value Theorem, there exists
a $\vartheta_h\in (0,1)$, such that
$$
\frac{f(a+h)-f(a)}{h}=f'(a+\vartheta_h h),
$$
and hence
$$
\left|\,\frac{f(a+h)-f(a)}{h}-L\,\right|=\left|\,f'(a+\vartheta_h h)-L\,\right|
<\varepsilon.
$$
The last inequality holds since $0<\vartheta_h h<h<\delta$.
A: Applying
L'Hospital's rule
yields: $
    \lim_{x \to a} \,(f(x) - f(a)) \mathbin{/} (x - a)
  = \lim_{x \to a} f'(x)
$.
Thus, the assumed existence of $\lim_{x \to a} f'(x)$
proves that the derivative of $f(x)$ at $x = a$ exists and is equal to
$\lim_{x \to a} f'(x)$.
A: Some hints:
Using the definition of derivative, you need to show that
$$
\lim_{h\rightarrow 0^+} {f(a+h)-f(a)\over h }
$$
exists and is equal to $\lim\limits_{x\rightarrow a^+} f'(x)$.
Note that for $h>0$ the Mean Value Theorem provides a point $c_h$ with $a<c_h<a+h$ such that
$$
{f(a+h)-f(a)\over h } =f'(c_h).
$$
Finally, note that $c_h\rightarrow a^+$ as $h\rightarrow0^+$.
A: A different version of the question: 

let $f:(a,b)\longrightarrow\mathbb{R}$ be a differentiable function in $(a,b)$ 
  except one point $x_{0}\in(a,b)$ where we don't know if $f$ is differentiable at.
let $lim_{x\rightarrow x_{0}}f'(x)=L\in\mathbb{R}$. 
?prove: $f$ is differentiable in $x=x_{0}$ and $f'(x_{0})=L$ .

according to the darbouex's therom, the derivative could only have a discontinuity point of the type “removable discontinuity”
where $lim_{x\longrightarrow x_{0}^{+}}\frac{f(x)-f(x_{0})}{x-x_{0}}\stackrel{\color{red}{**}}{\color{black}{\neq}} lim_{x\longrightarrow x_{0}^{-}}\frac{f(x)-f(x_{0})}{x-x_{0}}$ . therefore, I'll preform a contradiction proof. 
Let's suppose that $lim_{x\longrightarrow x_{0}^{+}}\frac{f(x)-f(x_{0})}{x-x_{0}}\neq lim_{x\longrightarrow x_{0}^{-}}\frac{f(x)-f(x_{0})}{x-x_{0}}$ so that $f$ is not difrrentiable at $x_{0}$. 
using the darbouex's therom, let $f'(a)$ be the derivative of $f$ to the right of $a$; 
$$lim_{x\rightarrow a^{+}}\frac{f(x)-f(a)}{x-a}= f'(a) , lim_{x\rightarrow b^{-}}\frac{f(x)-f(b)}{x-b} = f'(b)$$ and let $$ f'(x_{1}) = lim_{x\rightarrow x_{0}^{-}}\frac{f(x)-f(x_{0})}{x-b}  , f'(x_{2}) =  lim_{x\longrightarrow x_{0}^{+}}\frac{f(x)-f(x_{0})}{x-x_{0}}$$.
in the interval of $(a,x_{0})$ where $f$ is differentiable, for every $y_{1}$ as  $y_{1}  $ is between $f'(a)$ and $f'(x_{1}):$ there exist $t_{1}\in[a,x_{0}]:f'(t_{1})=y_{1}$. 
in the interval of $(x_{0},b)$ where $f$ is differentiable, for every $y_{2}$ as $y_{2}$ is between $f'(x_{2})$ and $f'(b): $ there exist $t_{2}\in[x_{0},b]:f'(t_{2})=y_{2}$. 
therefore, according to $\color{red}{**}$ $$lim_{t_{1}\longrightarrow x_{0}^{-}}f'(t_{1})\neq lim_{t_{2\longrightarrow}x_{0}^{+}}f'(t_{2})$$ but that's a contradiction to the fact that $$lim_{x\rightarrow x_{0}}f'(x)=L=lim_{t_{1}\longrightarrow x_{0}^{-}}f'(t_{1})\neq lim_{t_{2\longrightarrow}x_{0}^{+}}f'(t_{2})=lim_{x\rightarrow x_{0}}f'(x)=L $$ hence, $f$ is differentiable at $x_{0}$ , and $f'(x_{0})=L$ .
A: Comment
There is something wrong with your statement. Notice that you only make the assumption that $f$ is continuous over $[a,b]$ and differentiable over $(a,b)$. Hence，we can only study the right limit of $f'(x)$,namely $\lim\limits_{x \to a+}f'(x)$. Now, I post the proper statement and its proof as follows.
Theorem

Suppose that $f(x)$ is continuous over $[x_0,x_0+H](H>0)$,and has its
  limited derivative $f'(x)$ when $x>x_0$. If $\lim\limits_{x \to
> x_0+}f'(x)=K,$ then the right derivative of $f(x)$ at $x_0$ equals $K$
  as well.

Proof
According to $MVT$, when $0<\Delta x \leq H$, we have $$\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}=f'(c),\tag 1$$ where $x_0<c<x_0+\Delta x.$ Therefore, if $\Delta x \to 0+$, then $c \to x_0+.$ Thus, by taking the simultaneous limits of the both sides of $(1)$,we have $$ \lim_{\Delta x \to 0+}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}=\lim_{\Delta x\to 0+}f'(c)=\lim_{c\to x_0+}f'(c)=K.$$This is what we want to prove.
