TITLE: Show $\lim_{h \to \ 0} \frac{f(x + 2h) - 2f(x+h) + f(x)}{h^{2}} = f''(x)$ Proof verification
QUESTION [7 upvotes]: Show $$\lim_{h \to \ 0} \frac{f(x + 2h) - 2f(x+h) + f(x)}{h^{2}} = f''(x)$$ 
Proof:
By definition:  
$$f'(x) = \lim_{h \to \ 0} \frac{f(x + h) - f(x)}{h}$$
Using this idea it would imply: 
$$1)\ \ f''(x) = \lim_{h \to \ 0} \frac{f'(x + h) - f'(x)}{h}$$
As such it is required that I find an expression for $f'(x+h)$. This is where I'm not sure if the step I took is legitimate.
An expression for $f'(x + h)$ is:
$$f'(x+h) =  \lim_{h \to \ 0} \frac{f(x + 2h) - f(x + h)}{h}$$
Combining this with the definition of $f'(x)$ and inserting it into 1) you arrive at: $$\lim_{h \to \ 0} \frac{f(x + 2h) - 2f(x+h) + f(x)}{h^{2}} = f''(x)$$ 
As required.  
Concern: I feel a discomfort with this solution. Even though "mechanically" it worked out, if I am taking the limit as $h \rightarrow 0$ that would mean $x + 2h$ and $x + h$ both go to $x$. But I am attempting to use the idea that $x +2h$ goes to $x + h$. Perhaps it is a notation idea that I need to communicate better, but I feel it is larger than just that.

REPLY [5 votes]: Let $f$ be any continuous function which is second differentiable
at $a$ (i.e. $f''(a)$ exists). By definition of second differentiability at a point, $f'(x)$ exists over $(a-\eta,a+\eta)$ for some $\eta > 0$. Define $g : (-\eta, \eta) \to \mathbb{R}$ by
$$g(x) = f(a+x) - f(a) - f'(a) x - \frac12 f''(a) x^2$$
It is clear $g$ is continuous, second differentiable at $a$ and $g'(x)$ exists over $(-\eta,\eta)$.
Furthermore, $g(0) = g'(0) = g''(0) = 0$. What we want to show is equivalent to following statement
$$\lim_{h\to 0}\frac{g(2h) - 2g(h) + g(0)}{h^2} = 0 \tag{*1}$$
Since $f''(a)$ exists, $g''(0) = \lim_{h\to 0}\frac{g'(h)}{h}$ exists and vanishes. For any $\epsilon > 0$, there exists a $\delta \in ( 0,\eta)$ such that
$$\left|\frac{g'(x)}{x}\right| = \left|\frac{g'(x) - g'(0)}{x}\right| < \frac{\epsilon}{3}\quad\text{ whenever }\quad 0 < |x| < \delta$$
For any $h$ such that $0 < |h| < \frac{\delta}{2}$, apply MVT to $g(2h) - g(h)$ and $g(h) - g(0)$, we find there are $p,q \in (0,1)$ such that
$$\frac{g(2h) - g(h)}{h}  = g'((1+p)h)\quad\text{ and }\quad \frac{g(h) - g(0)}{h} = g'(qh)$$
This leads to
$$\begin{align}\left|\frac{g(2h) - 2g(h) +  g(0)}{h^2}\right| 
& \le \left| \frac{g(2h) - g(h)}{h^2}\right| + \left|\frac{g(h) - g(0)}{h^2}\right|\\
&= \left|\frac{g'((1+p)h)}{h}\right| + \left|\frac{g'(qh)}{h}\right| \\
& < \frac{\epsilon}{3}\left((1+p) + q\right)\\
& < \epsilon
\end{align}
$$
Since $\epsilon$ can be arbitrary small, $(*1)$ follows and hence
$$\lim_{h\to 0}\frac{f(a+2h) - 2f(a+h) + f(a)}{h^2} = f''(a)$$