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If $f$ is twice derivable at $a$, compute $$\lim_{h\to 0} \frac{f(a+h)+f(a-h)-2f(a)}{h^2}$$

I tried to rearrange the terms in this way $$\lim_{h\to0} \frac{f(a+h)-f(a)}{h^2} + \lim_{h\to 0}\frac{f(a-h)-f(a)}{h^2}$$

but it didn't get me anywhere. I also realised I can't use l'Hôpital rule or Taylor's theorems, because the hypothesis is weaker.

A. Salguero-Alarcón
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    Note that $\frac{f(a+h)-f(a)}{h^2} = \frac{\left(\frac{f(a+h)-f(a)}{h}\right)}{h}$ – joeb Feb 20 '17 at 23:13
  • http://math.stackexchange.com/questions/221905/check-my-workings-show-that-lim-h-to0-fracfxh-2fxfx-hh2-fx – kingW3 Feb 20 '17 at 23:15

2 Answers2

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I also realised I can't use l'Hôpital rule or Taylor's theorems, because the hypothesis is weaker.

The hypothesis is in fact sufficient to use L'Hôpital once.

The numerator has limit $\lim_{h\to 0} f(a+h)+f(a-h)-2f(a) = 2f(a)-2f(a)=0$ by continuity of $f$, and the denominator obviously tends to $0$ as well. Both are differentiable in $h\,$, then by L'Hôpital's rule the limits are equal provided that the RHS one exists:

$$\lim_{h\to 0} \frac{f(a+h)+f(a-h)-2f(a)}{h^2} = \lim_{h \to 0} \frac{f'(a+h)-f'(a-h)}{2h}$$

But the RHS limit is $f''(a)$ by the definition of derivatives, and given that $f'$ is differentiable at $a$ the limit exists, so the previous step holds, and the original limit exists and is equal to $f''(a)\,$.

dxiv
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I give a proof for $f$ of $C^2$ class.

It is simply $f''(a)$. Namely, the expression under limit equals to a doubled divided difference $[a-h,a,a+h;f]$. By a MVT for divided differences we infer that this difference equals to $\frac{f''(\xi)}{2}$ (so doubled difference is $f''(\xi)$) for some $\xi$ in the interval with endpoints $a-h,a+h$. If $h\to 0$, then $\xi\to a$. If $f$ is of $C^2$ class, then $f''(\xi)\to f''(a)$ when $h\to 0$.

szw1710
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