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.

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.
