Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 92

Prove that $\displaystyle f''(x) = \lim_{h \to 0 } \frac{f(x+1)-2f(x) + f(x-h)}{h^2}$ suppose that $f''(x)$ is continuous.

By applying L'Hospital's Rule
$\displaystyle f''(x) = \lim_{h \to 0 } \frac{f(x+1)-2f(x) + f(x-h)}{h^2} = \lim_{h \to 0} \frac{f'(x+h)(1)+f'(x-h)(-1)}{2h}$

Again, we must apply L'Hospital's Rule since the limit is an indeterminate form

$\begin{equation} \begin{aligned} \lim_{h \to 0} \frac{f'(x+h)(1)+f'(x-h)(-1)}{2h} &= \lim_{h \to 0} \frac{f''(x+h) - f''(x-h)(-1)}{2}\\ \\ &= \lim_{h \to 0} \frac{2f''(x+h)}{2}\\ \\ &= \lim_{h \to 0} f''(x+h)\\ \\ &= f''(x+0)\\ \\ &= f''(x) \end{aligned} \end{equation}$