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 first proof:
$$lim_{h\rightarrow0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2}=$$using De L'Hopital Rule
$$=lim_{h\rightarrow0}\frac{f'(a+h)+f'(a-h)}{2h}=\frac{f''(a+h)+f''(a-h)}{2}=\frac{f''(a)+f''(a)}{2}=f''(a)$$
Instead the second question:
$$\lim_{h\rightarrow0}\frac{4(f(h)-f(-h)-2(f(h/2)-f(-h/2)))}{h^3}=$$ applying De L'Hopital Rule
$$\lim_{h\rightarrow0}\frac{4(f'(h)+f'(-h)-f'(h/2)-f'(-h/2))}{3h^2} \\
= \lim_{h\rightarrow0}\frac{4(f''(h)-f''(-h)-\frac{1}{2}f''(h/2)+\frac{1}{2}f''(-h/2))}{6h} \\
= \lim_{h\rightarrow0}\frac{4(f'''(h)+f'''(-h)-\frac{1}{4}f'''(h/2)-\frac{1}{4}f'''(-h/2))}{6} \\
= \frac{6f'''(0)}{6} = f'''(0)$$
