Existence of second derivative implies symmetric second derivative I would like to know if the following statement is true.

Let $f:(a,b)\subset\mathbb{R}  \rightarrow \mathbb{R}$, such $f'$
  exits for all $x \in (a,b)$ and also exists $f''(x_0)$ for some
  $a<x_0<b$. Then
$ \lim_{h \to 0} \frac{f(x_0+h)-2f(x_0)+f(x_0-h)}{h^2} = f''(x_0)$.

Obs: I'm not assuming that $f'$ is continuos in some neighborhood of $x_0$. However, I think that requirement of the existence of $f'$ in a neighborhood of $x_0$ is a necessity for the existence of $f''(x_0)$.
 A: By Taylor's formula,
$$f(x_0 + h) = f(x_0) + hf'(x_0) + \frac{1}{2}h^2f''(x_0) + o(h^2)$$
and 
$$f(x_0 - h) = f(x_0) - hf'(x_0) + \frac{1}{2}h^2f''(x_0) + o(h^2)$$
as $h \to 0$.
This shows that the expression whose limit you are to compute is equal to $f''(x_0) + o(1)$.
A: Let $\varphi(t)=f(x_{0}+th)-2f(x_{0})-f(x_{0}-(1-t)h)$, $0\leq t\leq 1$, then $\varphi(1)-\varphi(0)=\varphi'(c_{h})$ for some $0<c_{h}<1$, so $\varphi(1)-\varphi(0)=[f'(x_{0}+c_{h}h)-f'(x_{0}-(1-c_{h})h)]h$. For $\epsilon>0$, find some $\delta>0$ such that if $0<|u|<\delta$,
\begin{align*}
\left|\frac{f'(x_{0}+u)-f'(x_{0})}{u}-f''(x_{0})\right|<\epsilon/2.
\end{align*}
Now for $0<|h|<\delta$,
\begin{align*}
\left|\frac{1}{h^{2}}[\varphi(1)-\varphi(0)]-f''(x_{0})\right|&\leq\left|\frac{f'(x_{0}+c_{h}h)-f'(x_{0})}{h}-c_{h}f''(x_{0})\right|\\
&~~~~+\left|\frac{f'(x_{0}-(1-c_{h})h)-f'(x_{0})}{-h}-(1-c_{h})f''(x_{0})\right|\\
&\leq|c_{h}|\left|\frac{f'(x_{0}+c_{h}h)-f'(x_{0})}{c_{h}{h}}-f''(x_{0})\right|\\
&~~~~+|1-c_{h}|\left|\frac{f'(x_{0}-(1-c_{h})h)-f'(x_{0})}{-(1-c_{h})h}-f''(x_{0})\right|\\
&<\epsilon.
\end{align*}
