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 $$\lim_{h\rightarrow0}\frac{4(f(h)-f(-h)-2(f(h/2)-f(-h/2)))}{h^3}=f'''(0)$$
 A: Question 1
$$
\lim_{h\rightarrow0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2} \\
$$
when $h = 0$ is substituted numerator and denominator reduce to $0$. So, applying L'Hopital's rule (differentiate wrt $h$)
$$
\lim_{h\rightarrow0}\frac{f'(a+h)-f'(a-h)}{2h} \\
\lim_{h\rightarrow0}\frac{f'(a+h)-f'(a)-f'(a-h)+f'(a)}{2h} \\
\lim_{h\rightarrow0}\frac{f'(a+h)-f'(a)}{2h}+\lim_{h\rightarrow0}\frac{f'(a)-f'(a-h)}{2h} \\
= \frac{2f''(a)}{2} = f''(a)\\
$$
Question 2
$$
\lim_{h\rightarrow0}\frac{4(f(h)-f(-h)-2(f(h/2)-f(-h/2)))}{h^3}
$$
when h=0 is substituted numerator and denominator reduce to 0. So, applying L'Hopital's rule (differentiate wrt h)
$$
\lim_{h\rightarrow0}\frac{4(f'(h)+f'(-h)-f'(h/2)-f'(-h/2))}{3h^2} \\
= \lim_{h\rightarrow0}\frac{4(f''(h)-f''(0)+f''(0)-f''(-h)+\frac{1}{2}f''(0)-\frac{1}{2}f''(h/2)-\frac{1}{2}f''(0)+\frac{1}{2}f''(-h/2))}{6h} \\
= \lim_{h\rightarrow0}\frac{4(f''(h)-f''(0)+f''(0)-f''(-h)+\frac{1}{2}f''(0)-\frac{1}{2}f''(h/2)-\frac{1}{2}f''(0)+\frac{1}{2}f''(-h/2))}{6h} \\
= \frac{2}{3}\left(\lim_{h\rightarrow0}\frac{f''(h)-f''(0)}{h}+\lim_{h\rightarrow0}\frac{f''(0)-f''(-h)}{h}\right)+\frac{1}{3}\left(\lim_{h\rightarrow0}\frac{f''(0)-f''(h/2)}{h}+\lim_{h\rightarrow0}\frac{f''(-h/2)-f''(0)}{h}\right) \\
= \frac{4f'''(0)}{3}-\frac{f'''(0)}{3} = f'''(0)
$$
PS: Thanks to @Paramanand Singh for pointing out that f was not twice and thrice differentiable in first and second questions respectively.
A: 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: $$\begin{aligned}L &= \lim_{h \to 0}\frac{4\left\{f(h) - f(-h) - 2\{f(h/2) - f(-h/2)\}\right\}}{h^{3}}\\
&= \lim_{h \to 0}\frac{4\left\{f'(h) + f'(-h) - \{f'(h/2) + f'(-h/2)\}\right\}}{3h^{2}}\text{ (by L'Hospital's Rule)}\\
&= \lim_{h \to 0}\frac{4\left\{f''(h) - f''(-h) - (1/2)\{f''(h/2) - f''(-h/2)\}\right\}}{6h}\text{ (by L'Hospital's Rule)}\\
&= \lim_{h \to 0}\frac{4\left\{f''(h) - f''(0) + f''(0) - f''(-h) - (1/2)\{f''(h/2) - f''(0) + f''(0) - f''(-h/2)\}\right\}}{6h}\\
&= \frac{2}{3}\lim_{h \to 0}\frac{f''(h) - f''(0)}{h} + \frac{2}{3}\lim_{h \to 0}\frac{f''(-h) - f''(0)}{-h} - \frac{1}{6}\lim_{h \to 0}\frac{f''(h/2) - f''(0)}{h/2}\\
&\,\,\,\,\,\,\,-\frac{1}{6}\lim_{h \to 0}\frac{f''(-h/2) - f''(0)}{-h/2}\\
&= \frac{2}{3}f'''(0) + \frac{2}{3}f'''(0) - \frac{1}{6}f'''(0) - \frac{1}{6}f'''(0)\\
&= f'''(0)\end{aligned}$$
A: 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
$$\lim_{h\rightarrow0}\frac{4(f'(h)+f'(-h)-f'(h/2)-f'(-h/2))}{3h^2} \\
= \lim_{h\rightarrow0}\frac{4(f''(h)-f''(-h)-\frac{1}{2}f''(h/2)+\frac{1}{2}f''(-h/2))}{6h} \\
= \lim_{h\rightarrow0}\frac{4(f'''(h)+f'''(-h)-\frac{1}{4}f'''(h/2)-\frac{1}{4}f'''(-h/2))}{6} \\
= \frac{6f'''(0)}{6} = f'''(0)$$
