Prove that $\lim_{h\to 0} \frac{g(a+h)-2g(a)+g(a-h)}{h^2} = g''(a)$ How do I prove that
$$\displaystyle\lim_{h\to0} \dfrac{g(a+h)-2g(a)+g(a-h)}{h^2} = g''(a),$$
where $g$ is of class $C^2$?
My attempt:
By Mean Value Theorem, there are $c_0\in (a-h, a)$ and $c_1\in(a, a+h)$ such that 
$$g(a)-g(a-h) = g'(c_0)\cdot  h$$
$$g(a+h)-g(a) = g'(c_1)\cdot  h$$
Then
$\displaystyle\lim_{h\to0} \dfrac{g(a+h)-2g(a)+g(a-h)}{h^2} = \displaystyle\lim_{h\to0} \dfrac{g(a+h)-g(a)-(g(a)-g(a-h))}{h^2} = \\ 
\displaystyle\lim_{h\to0} \dfrac{g'(c_1)-g'(c_0)}{h}=
\displaystyle\lim_{h\to0} \dfrac{g'(c_1)-g'(c_0)}{c_1-c_0}\cdot\dfrac{c_1-c_0}{h}=\\
\displaystyle\lim_{h\to0} \dfrac{g'(c_1)-g'(c_0)}{c_1-c_0}\cdot\displaystyle\lim_{h\to0}\dfrac{c_1-c_0}
{h}
$
The first limit is equals to $g''(a)$. Then the second limit should be equals to $1$, but I can't see why.
 A: If we know $g''(x)$ exists for an open interval $I$ containing $a$, $$\lim_{h\to0} \frac{g(a+h)-2g(a)+g(a-h)}{h^2}=\lim_{h\to0} \frac{g'(a+h)-g'(a-h)}{2h}
=\displaystyle\lim_{h\to0} \frac{g'(a+h)-g'(a)+g'(a)-g'(a-h)}{2h} =g''(a)
$$, by L'Hopital Rule
A: By Taylor's expansion


*

*$g(a+h)=g(a)+g'(a)h+\frac12g''(a)h^2+o(h^2)$

*$g(a-h)=g(a)-g'(a)h+\frac12g''(a)h^2+o(h^2)$


and adding up
$$g''(a)=\frac{g(a+h)-2g(a)+g(a-h)}{h^2}+\frac{o(h^2)}{h^2}$$
then take the limit.
A: \begin{align*}
g(a+h)&=g(a)+g'(a)h+\dfrac{1}{2}g''(\xi_{h})h^{2}\\
g(a-h)&=g(a)-g'(a)h+\dfrac{1}{2}g''(\eta_{h})h^{2},
\end{align*}
where $\xi_{h}$ lies in between $a$ and $a+h$, and $\eta_{h}$ lies in between $a$ and $a-h$, so 
\begin{align*}
\dfrac{1}{h^{2}}[g(a+h)+g(x-h)-2g(a)]=\dfrac{1}{2}(g''(\xi_{h})+g''(\eta_{h}))\rightarrow\dfrac{1}{2}(g''(a)+g''(a))=g''(a)
\end{align*}
by the continuity of $g''$.
A: Note that
$$\dfrac{g(a+h)-2g(a)+g(a-h)}{h^2} =\dfrac{\dfrac{g(a+h)-g(a)}{h} -\dfrac{g(a)-g(a-h)}{h}}{h},$$
so that quotient is basically the slope of $g'$ at $x=a$ for $h$ very small.
