Use L'Hopital's Rule to Prove Let $$f: \mathbb R\rightarrow \mathbb R$$ be differentiable, let a in $\mathbb R$. Suppose that $f''(a)$ exists. Prove that $$\lim_{h\rightarrow0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2}=f''(a) $$
Suppose further that $f''(x)$ exists for all $x$, and that $f'''(0)$ exists. Prove that $$\lim_{h\rightarrow0}\frac{4(f(h)-f(-h)-2(f(h/2)-f(-h/2)))}{h^3}=f'''(0)$$

The earlier accepted answer here (by Priyatham) had a subtle flaw (namely repeated use of L'Hospital's Rule without checking conditions of its applicability) but it has been fixed now after my comments.
I am sure this question is solved elsewhere on MSE, but I am not able to find link. Here is the right approach to this question.
1) We will use L'Hospital's rule as follows: $$\begin{aligned}L &= \lim_{h \to 0}\frac{f(a + h) - 2f(a) + f(a - h)}{h^{2}}\\
&= \lim_{h \to 0}\frac{f'(a + h) - f'(a - h)}{2h}\text{ (by L'Hospital's Rule)}\\
&= \lim_{h \to 0}\frac{f'(a + h) - f'(a) + f'(a) - f'(a - h)}{2h}\\
&= \lim_{h \to 0}\frac{f'(a + h) - f'(a)}{2h} + \lim_{h \to 0}\frac{f'(a - h) - f'(a)}{-2h}\\
&= \frac{f''(a)}{2} + \frac{f''(a)}{2} = f''(a)\end{aligned}$$ 2) Again we use L'Hospital's rule: $$\begin{aligned}L &= \lim_{h \to 0}\frac{4\left\{f(h) - f(-h) - 2\{f(h/2) - f(-h/2)\}\right\}}{h^{3}}\\
&= \lim_{h \to 0}\frac{4\left\{f'(h) + f'(-h) - \{f'(h/2) + f'(-h/2)\}\right\}}{3h^{2}}\text{ (by L'Hospital's Rule)}\\
&= \lim_{h \to 0}\frac{4\left\{f''(h) - f''(-h) - (1/2)\{f''(h/2) - f''(-h/2)\}\right\}}{6h}\text{ (by L'Hospital's Rule)}\\
&= \lim_{h \to 0}\frac{4\left\{f''(h) - f''(0) + f''(0) - f''(-h) - (1/2)\{f''(h/2) - f''(0) + f''(0) - f''(-h/2)\}\right\}}{6h}\\
&= \frac{2}{3}\lim_{h \to 0}\frac{f''(h) - f''(0)}{h} + \frac{2}{3}\lim_{h \to 0}\frac{f''(-h) - f''(0)}{-h} - \frac{1}{6}\lim_{h \to 0}\frac{f''(h/2) - f''(0)}{h/2}\\
&\,\,\,\,\,\,\,-\frac{1}{6}\lim_{h \to 0}\frac{f''(-h/2) - f''(0)}{-h/2}\\
&= \frac{2}{3}f'''(0) + \frac{2}{3}f'''(0) - \frac{1}{6}f'''(0) - \frac{1}{6}f'''(0)\\
&= f'''(0)\end{aligned}$$
