Using L'Hospital's rule to prove $f'(x)= \lim_{h \to 0} \frac {f(x+h)- f(x-h)}{2h}$ $$
f'(x)= \lim_{h \to 0} \frac {f(x+h)- f(x-h)}{2h}. $$ It is given that the derivative of $f$ at $x$ exists. 
If I use L'Hospital's rule and differentiate top and bottom wrt $h$, I get
$$\lim_{h \to 0} \frac {f'(x+h) + f'(x-h)} {2}$$
Then what? I am really lost on this one. I can get the result without using the rule, but the challenge is to use it. Any ideas?
 A: L'Hospital is the wrong approach since you only know that $f$ is differentiable at $x$, and you don't know that $f'$ is continuous. That is what you need for that approach.
Anyhow
$$ \lim_{h \to 0} \frac {f(x+h)- f(x-h)}{2h}=\frac{1}{2} \left[  \lim_{h \to 0} \frac {f(x+h)- f(x)}{h}+ \lim_{h \to 0} \frac {f(x)- f(x-h)}{h}\right]$$
By the definition of differentiability at $x$ we have
$$ \lim_{h \to 0} \frac {f(x+h)- f(x)}{h}=f'(x)$$
and after the substitution $h=-t$ you get
$$\lim_{h \to 0} \frac {f(x)- f(x-h)}{h}=\lim_{t \to 0} \frac {f(x)- f(x+t)}{-t}
=\lim_{t \to 0} \frac {f(x+t)- f(x)}{t} =f'(x)$$
thus
$$ \lim_{h \to 0} \frac {f(x+h)- f(x-h)}{2h}=\frac{1}{2} [ f'(x)+f'(x)]=f'(x) \,.$$
P.S. If the problem asks you explicitly to use L'H, and you don't have the extra condition that $f$ is differentiable around $x$ and $f'$ is continuous at $x$, then the problem is wrong.
Let $f(x)=x^2\sin(\frac{1}{x})$ with $f(0)=0$. Then by the Squeze Theorem you can prove that $f'(0)=0$.
But 
$$f'(x)=2x \sin(\frac{1}{x})-\cos(\frac{1}{x})$$
and hence
$$\lim_{h \to 0} \frac {f'(0+h)+ f'(0-h)}{2}=\lim_{h \to 0} \frac {2h \sin(\frac{1}{h})-\cos(\frac{1}{h})+2h \sin(\frac{1}{h})-\cos(\frac{1}{h})}{2}=\mbox{Does Not Exist}$$
A: $$f'(x)= \lim_{h \to 0} \frac {f(x+h)- f(x-h)}{2h}=\lim_{h \to 0} \frac {f'(x+h) + f'(x-h)} {2} = \frac{2f'(x)}{2}=f'(x)$$
though it's a bit circular.
Your mistake was when you differentiated, you forgot to change the sign before $f'(x-h)$.
A: The question is (implicitly) assuming that the derivative of $f$ is continuous at $x$. Note that you forgot the inner derivative of the second term in the numerator, so that the limit you got is not correct. The correct limit you get is $\lim_{h\to 0} (f'(x+h)+f'(x-h))/2$ which, by continuity of $f'$ at $x$, evaluates to $(f'(x)+f'(x))/2=f'(x)$.
