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

\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*}

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 $h\to 0$.

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