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?
Best Answer
\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*}