TITLE: Show that $f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h}$
QUESTION [11 upvotes]: I have to prove that if a function $f$ is differentiable on $(a,b)$, then
\begin{align*}
f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h}
\end{align*}
Using the fact that $f'(x) = \lim\limits_{h \rightarrow 0}\dfrac{f(x+h)-f(x)}{h}$, I wrote my proof in the following manner:
\begin{align*}
\lim\limits_{h \rightarrow 0}f(x+h) - 2f(x) = \lim\limits_{h \rightarrow 0} - f(x-h)
\\
\lim\limits_{h \rightarrow 0} \dfrac{f(x+h)}{2h} - \dfrac{f(x)}{h} = \lim\limits_{h \rightarrow 0}\dfrac{-f(x-h)}{2h}
\\
\lim\limits_{h \rightarrow 0}\dfrac{f(x+h)}{h} - \dfrac{f(x)}{h} = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)}{2h} - \dfrac{f(x-h)}{2h}
\\
\lim\limits_{h \rightarrow 0} \dfrac{f(x+h) - f(x)}{h} = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h}
\end{align*}
However, I believe that it is actually incorrect, because when I divide by $2h$ I am potentially making the limit undefined. How would I go about correcting my proof?

REPLY [3 votes]: Since $f$ is differentiable, we may use l'Hopitals rule:
$$\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}=\lim_{h\to 0}\frac{f'(x+h)-(-1)f'(x-h)}{2}=\lim_{h\to 0}\frac{f'(x+h)+f'(x-h)}{2}=\frac{2f'(x)}{2}=f'(x).$$