Proving Derivative at 0 exists I'm really struggling on this question.  I've thought about it for a long time now but I can't seem to figure it out.  I know I am supposed to use the Mean Value Theorem in $[0,h]$ with $h>0$ and use the Mean Value Theorem in $[h,0]$ with $h<0$. 
Attempt: I applied the definition of mean value Theorem for both of these and get:
There exists c such that $f'(c)=f(h)-f(0)/h$.  There exists d such that $f'(d)=f(0)-f(h)/-h$.  $f'(c)$ clearly equals $f'(d)$, which implies c=d=0.  Thus f'(0) exists.  I'm not sure if this is correct, but how would I conclude it equals L?
I know that there is some theorem that $f\in C[a,b],  $  f is differentiable in [a,b).  Then if $\lim\limits_{x \to a+} \ f'(x)\in R$, then $L=f'(a)$.  
I think then the fact given in the question that $\lim\limits_{x \to 0} \ f'(x)\ =L$, would imply that $f'(0)$ equals L from this theorem, and hence we are finished.
Is this correct?  I'm not sure if I did it correctly, but have given my best attempt.  Any help would be much welcomed.  
 A: For the interval $[0,h]$, the MVT states that there is some $x \in (0,h)$ such that
$$
f'(x) = \frac{f(h)-f(0)}{h}
$$
Let $h \downarrow 0$. Then $x \downarrow 0$ because it's in $(0,h)$ and, and we have
$$
\lim_{x \downarrow 0} \ f'(x) = f_+'(0)
$$
where the RHS follows from the definition of a right-hand derivative. Also, the LHS is equal to $L$, as it was given. Thus $f'_+(0) = L$.
A similar argument for the interval $[-h,0]$ yields $f_-'(0) = L$, from which it follows that $L = f'_+(0) = f_-'(0) = f'(0)$.
A: The part : $f'(c)=f'(d) \implies c=d=0$ is not very clear.
Consider for example $f(x)=x^3$, then $f'(x)=f'(-x)$ for any $x \in \mathbb{R}$. That does not imply $x=-x$.
In addition, avoid talking about differentiability on closed or semi-closed intervals. $f'(a)$ is not well defined on $[a,b)$. I am not saying that the theorem you are talking about does not exist at all, but it probably does not exist the way you are describing it. 
The proof given by Ken Wei is the one you are supposed to do, according to the hint.
