Proof that $\frac{f(x+h)-2 \cdot f(x)+f(x-h)}{h^2}\to f''(x)$ I want to proof that if $f \in C^2(\mathbb{R})$ then, for every $x \in \mathbb{R}$, $$\lim\limits_{h \to 0} \frac{f(x+h)-2 \cdot f(x)+f(x-h)}{h^2}=f''(x).$$
My attempt:
$$\begin{aligned}
f''(x)&= \lim\limits_{h \to 0} \frac{f'(x)-f'(x-h)}{h}\\\\
&= \lim\limits_{h \to 0} \Bigg(\frac{\frac{f(x+h)-f(x)}{h}-\frac{f(x)-f(x-h)}{h}}{h}\Bigg)\\\\
&= \lim\limits_{h \to 0} \frac{\frac{f(x+h)-2 \cdot f(x)+f(x-h)}{h}}{h}\\\\
&= \lim\limits_{h \to 0} \frac{f(x+h)-2 \cdot f(x)+f(x-h)}{h^2}
\end{aligned}$$
Now I have a few questions regarding my solution:

*

*Is my attempt correct for $f \in C^2(\mathbb{R})$ ? 

*I don't see that I have used that $f \in C^2(\mathbb{R})$, only the part of it that tells me that $f$ is differentiable twice in $x \in \mathbb{R}$. Am I right, or did I use the fact that $f$ is continuously differentiable twice without realizing ?

*If I have not used the fact that $f$ is continuously differentiable twice, wouldn't that mean that the proof is also valid if $f$ is not necessarily $C^2(\mathbb{R})$ but twice differentiable in $x \in \mathbb{R}$ ? If not, how can I prove it for $f$ twice differentiable but not necessarily $C^2(\mathbb{R})$ ?

 A: By L'Hopital's rule (differentiating $h \mapsto f(x+h)$ and $h \mapsto f(x-h)$ in the first step)
$$\lim_{h \to 0}\frac{f(x+h)-2 \cdot f(x)+f(x-h)}{h^2}= \lim_{h \to 0}\frac{f'(x+h) - f'(x-h)}{2h} \\= \frac{1}{2} \left[\lim_{h \to 0}\frac{f'(x+h) - f'(x)}{h}+ \lim_{h \to 0}\frac{f'(x) - f'(x-h)}{h}\right]= f''(x)$$
A: Your approach is problematic: when you replace $f'(x)$ and $f'(x-h)$ with difference quotients, the corresponding limits must be independent of $h$. In particular, you don't even get the following:
$$
\lim_{h\to0}\frac{f'(x) - f'(x-h)}{h} = \lim_{h\to0} \frac{f'(x) - \frac{f(x)-f(x-h)}{h}}{h}
$$
without some additional work.
The most concise proof I know of the statement involves Taylor series, namely the Lagrange Remainder Theorem. Let $f \in C^2(\Bbb{R})$ be given, and choose any $x \in \Bbb{R}$. Then, by LRT, for any $h > 0$ there exist $\xi_h \in (x,x+h)$ and $\eta_h \in (x-h,x)$ such that
$$
f(x+h) = f(x) + hf'(x) + \frac{h^2}{2}f''(\xi_h) \text{ and } f(x-h) = f(x) - hf'(x) + \frac{h^2}{2}f''(\eta_h).
$$
Add these two equations and subtract $2f(x)$ to reveal
$$
f(x+h) - 2f(x) + f(x-h) = \frac{h^2}{2}\left(f''(\xi_h) + f''(\eta_h)\right).
$$
Dividing by $h^2$ gives you your difference quotient on the LHS. Hence to complete the proof, we only have to show that $\lim_{h\to0} \frac{1}{2}\left(f''(\xi_h) + f''(\eta_h)\right) = f''(x)$.
See if you can complete the argument from there. Hint: Show $\lim_{h\to0} \xi_h = \lim_{h\to0} \eta_h = x$ and use continuity of $f''$ at $x$.
A: Your approach is right to me, even quite rough.
If you can use Taylor expansion, I'd suggest to follow that way, i.e. expanding $f(x+h)$ and $f(x-h)$, and then performing $\frac{(f(x+h) - 2f(x) +f(x-h))}{h^2}$.
You use the double differentiability when you compute the derivative of the first derivative with the limit $h\rightarrow 0$.
In the first line of your question you forgot either a $\lim_{h\rightarrow 0}$ or a $o(h)$
