Show that $f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h}$

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?

Solution 1:

\begin{align*} f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \end{align*} but also \begin{align*} f'(x) = \lim\limits_{h \rightarrow 0} \dfrac{f(x)-f(x-h)}{h} \end{align*} sum them up and divide by 2 to get

\begin{align*} f'(x) = \lim\limits_{h \rightarrow 0}\frac{ \dfrac{f(x+h)-f(x)}{h} + \dfrac{f(x)-f(x-h)}{h}}2 =\lim\limits_{h \rightarrow 0} \dfrac{f(x+h)-f(x-h)}{2h} \end{align*}

Solution 2:

$$ f(x+h) = f(x)+f'(x)h + o(|h|) $$ together with: $$ f(x-h) = f(x)-f'(x)h + o(|h|) $$ gives: $$ f(x+h)-f(x-h) = 2h\cdot f'(x) + o(|h|).$$

Solution 3:

You cannot just divide by $h$ from nowhere, it is not correct. Consider splitting $$ \frac{f(x+h)-f(x-h)}{2h}=\frac{f(x+h)-f(x)+f(x)-f(x-h)}{2h}= \frac12\left(\frac{f(x+h)-f(x)}{h}+\frac{f(x-h)-f(x)}{-h}\right). $$ Both terms go to $f'(x)$ as $h\to 0$.