# Proving $\displaystyle\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)$

The function $f$ is differentiable twice at x. Prove that: $$\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)$$ Hint: use Peano's remainder (if $f:I\to\mathbb R$ is differentiable $n$ times on $a\in I$ then $R_k(x)=o(|x-a|^k), \ x\to a$).

I just don't see the connection here between the second derivative and the Taylor series which has to do with Peano's remainder...

The remainder is defined to be the difference between the function and polynomial but what does it has to do in this case ?

• Hint: $f(x+h) = f(x) + h.f'(x) + R_1(h)$ – Thomas Jan 11 '14 at 17:45

## 1 Answer

We write the Taylor series:

$$f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(x)+h^2\epsilon(h)\tag{1}$$ and $$f(x-h)=f(x)-hf'(x)+\frac{h^2}{2}f''(x)+h^2\epsilon'(h)\tag{2}$$ where $$\lim_{h\to0}\epsilon (h)=\lim_{h\to0}\epsilon' (h)=0$$ so add the two equalities $(1)$ and $(2)$ and you find the result easily.

• Thanks I see everything got canceled out but I'm going to need some explanation. Bear with me because we've just learned Taylor. That epsilon is the remainder I assume ? How did you know how to build the two Taylor series the way you did ? Why does the limit of the two epsilon go to zero ? – GinKin Jan 11 '14 at 18:02
• Just I applied the Taylor series: $$f(x+h)=f(x)+\frac{h}{1!}f'(x)+\frac{h^2}{2!}f''(x)+\cdots+\frac{h^n}{n!}f^{(n)}(x)+h^n\epsilon(h)$$ where $\epsilon$ is a function tends to zero when $h$ tends to zero. – user63181 Jan 11 '14 at 18:36
• Why does it tends to zero ? – GinKin Jan 12 '14 at 5:46
• This is a part of the statement of the Taylor theorem and of course to understand why you should read its proof. – user63181 Jan 12 '14 at 5:53
• @SamiBenRomdhane: I am not sure if the OP aware of the form of Taylor's theorem you have used. Most common textbooks don't use the form of remainder you have used. The form you have used makes the least amount of assumption on $f$, namely the existence of $f^{(n)}$ at a single point $x$. To OP, one proof of this form of Taylor's theorem is easily done by first expressing $\epsilon(h)$ in terms of $f(x + h), f(x), f'(x), \ldots$ and then applying LHR $(n - 1)$ times. – Paramanand Singh Jan 12 '14 at 7:09