How to understand this derivation. I don't understand how the first equality comes about. I read that L'Hopital's Rule was used.
$ \lim_{h \to 0} \frac{f(x + h) + f(x - h) - 2f(x)}{h^2}
 = \lim_{h \to 0} \frac{f'(x + h) - f'(x - h)}{2h} \\
 = f''(x).$
 A: $$\frac{\partial f(x+h)}{\partial h}=f'(x+h)$$
$$\frac{\partial f(x-h)}{\partial h}=-f'(x-h)$$
$$\frac{\partial f(x)}{\partial h}=0$$
$$(h^2)'=2h$$
A: i will use the fundamental theorem of calculus and integration by parts to derive 
$\lim_{h \to 0} \frac{f(x + h) + f(x - h) - 2f(x)}{h^2}= f''(x)$
the left hand side limit may exist even for a function that is twice differentiable. take for example $f(x) = |x|, x = 0.$ the left hand side is $2$ and right hand side does not exist.
we will assume that $f$ is three times differentiable.
$\begin{align}
f(x+h) - 2f(x) + f(x-h) &= 
\int_x^{x+h}f^\prime(t) dt - \int_{x-h}^{x}f^\prime(t) dt\\
&= (t - x - h)f^\prime(t) \rvert_x^{x+h} - \int_x^{x+h} (t - x - h) f^{\prime \prime}(t) dt \\
& \ \ - (t - x + h)f^\prime(t)\rvert_{x-h}^x + \int_x^{x+h} (t - x + h) f^{\prime \prime}(t) dt \\
& =- \int_x^{x+h} (t - x - h) f^{\prime \prime}(t) dt + \int_x^{x+h} (t - x + h) f^{\prime \prime}(t) dt \\
&= -\frac{1}{2}(t-x-h)^2 f^{\prime \prime}(t) \rvert_x^{x+h} + \frac{1}{2}(t-x-h)^2 f^{\prime \prime}(t) \rvert_{x-h}^{x} \\
& \ \ + \int_x^{x+h}\frac{1}{2}(t-x-h)^2 f^{\prime \prime \prime}(t) dt -
\int_x^{x+h}\frac{1}{2}(t-x-h)^2 f^{\prime \prime \prime}(t) dt\\
&=h^2f^{\prime \prime}(x)+Kh^3/6 \text{ where $K = f^{\prime \prime \prime}(c)$ for some $x-h < c < x + h$} 
\end{align}$
moving things around and taking the limit gives you the desired result.
A: It is very simple if you apply Taylor's series with Peano's residue:
When $h\to 0$, we have
$$f(x+h)=f(x)+hf'(x)+\frac{1}{2}h^2 f''(x)+o(h^2)$$
$$f(x-h)=f(x)-hf'(x)+\frac{1}{2}h^2 f''(x)+o(h^2)$$
where $o(h^2)$ denotes any infinitesimal amount that converges to zero faster than $h^2$.
Put them back to the limit, the result should be
$$f''(x)+\frac{o(h^2)}{h^2}$$
and as $h\to 0$, it is obvious that the residue $\frac{o(h^2)}{h^2} \to 0$, thus the final result should be 
$$f''(x)$$
