Answer
$\lim\limits_{h \to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2} = f''(x)$
Work Step by Step
$\lim\limits_{h \to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2} = \frac{f(x)-2f(x)+f(x)}{0} = \frac{0}{0}$
We can apply L'Hospital's Rule.
$\lim\limits_{h \to 0}\frac{f'(x+h)-2f'(x)+f'(x-h)}{2h} =\frac{f'(x)-2f'(x)+f'(x)}{0} = \frac{0}{0}$
Note that:
$\lim\limits_{h \to 0}\frac{f'(x+h)-2f'(x)+f'(x-h)}{2h} = \lim\limits_{h \to 0}\frac{f'(x+h)}{2h} +\lim\limits_{h \to 0}\frac{-2f'(x)+f'(x+h)}{2(-h)}$
We can apply L'Hospital's Rule.
$\lim\limits_{h \to 0}\frac{f''(x+h)}{2} +\lim\limits_{h \to 0}\frac{-2f''(x)+f''(x+h)}{-2}$
$= \frac{f''(x)}{2} -\frac{-2f''(x)+f''(x)}{2}$
$ = \frac{2f''(x)}{2}$
$ = f''(x)$
Therefore:
$\lim\limits_{h \to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2} = f''(x)$
