$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} $
now let $\displaystyle u = x+h \Rightarrow x = u-h $
So $\displaystyle f'(u-h) = \lim_{h \to 0} \frac{f(u-h+h)-f(u-h)}{h} = \lim_{h \to 0} \frac{f(u)-f(u-h)}{h} $
Since $\displaystyle h \to 0, f'(u-h) \to f'(u) $
Therefore $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x)-f(x-h)}{h} $
$\displaystyle f'(x)+f'(x) = 2f'(x) = \lim_{h \to 0}\left( \frac{f(x+h)-f(x)}{h} + \frac{f(x)-f(x-h)}{h} \right)$
$\displaystyle 2f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{h} $
Hence, $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{2h} $