TITLE: Proving that $\lim\limits_{h\to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)$
QUESTION [7 upvotes]: Prove that $$\lim\limits_{h\to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)$$

Is the following a correct proof:

$f''(x)=$$\lim\limits_{h\to 0}\frac{f'(x)-f'(x-h)}{h}=\lim\limits_{h\to 0}\frac{\frac{f(x+h)-f(x)}{h}-\frac{f(x)-f(x-h)}{h}}{h}=\lim\limits_{h\to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}$

I would really love input on this proof. The book "Berkeley Problems in Mathematics" solves it differently.

REPLY [2 votes]: Simply use L'Hospital rule twice
$\lim\limits_{h\to 0}\dfrac{f(x+h)-2f(x)+f(x-h)}{h^2}\\=\lim\limits_{h\to 0}\dfrac{f'(x+h)-f'(x-h)}{2h}\\=\lim\limits_{h\to 0}\dfrac{f''(x+h)+f''(x-h)}{2}\\=f''(x)$