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

$$f(x+h)=f(x)+hf'(x)+\frac{h^2}{2}f''(x)+h^2\epsilon(h)\tag{1}$$ 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.
• 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
• @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