TITLE: Can the limit $\lim_{h \to 0}\frac{f(x + h) - 2f(x) + f(x - h)}{h^2}$ exist if $f'(x)$ does not exist at $x$?
QUESTION [6 upvotes]: The second derivative of $f$ can be written as 
$$f''(x) = \lim_{h \to 0}\frac{f(x + h) - 2f(x) + f(x - h)}{h^2}$$
while it can also be written as (in fact, I believe this is the definition of $f''(x)$):
$$f''(x) = \lim_{h \to 0}\frac{f'(x + h) - f'(x)}{h}$$
From the second expression, it seems clear that if $f'(x)$ does not exist, then $f''(x)$ cannot exist as well.
However, the first expression does not involve $f'(x)$, so this got me wondering: If $f'(x)$ does not exist at $x$ , can the limit $\lim_{h \to 0}\frac{f(x + h) - 2f(x) + f(x - h)}{h^2}$ at $x$  exist?
How would one prove/disprove this? If this were to be true, what would be some examples?

REPLY [6 votes]: An easy counterexample is the case of an odd function that's not differentiable at $0$: let $f$ be any odd function (defined at $0$). Then:
$$\lim_{h\to0}\frac{f(0-h)-2f(0)+f(0+h)}{h^2}=\lim_{h\to0}\frac{f(-h)+f(h)}{h^2}=0.$$
So there are lots of counterexample, e.g., $f(x)=x^{1/3}$.