# Let $f$ a real function defined on a neighbourhood of a point $x$ and in the form $f''(x)=(f')'(x)$. Show that:

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

Please, i don't know where to even begin. How do i start? Please, give me a detailed hint, because i'm kind of lost in this class...

• See here and here. – Joey Zou Sep 25 '16 at 18:54

An alternative is to use Taylor's formula

$$f(x+h)=f(x)+hf'(x)+\dfrac{h^2}{2}f''(x)+o(h^2)$$ and $$f(x-h)=f(x)-hf'(x)+\dfrac{h^2}{2}f''(x)+o(h^2)$$

You can start with the definition of derivative: $$f''(x) = (f')'(x) = \lim_{h\rightarrow 0}\frac{f'(x+h)-f'(x)}{h}$$

• but i'll put a limit behind a limit? – Lucas Barbiere Sep 25 '16 at 18:30
• @LucasBarbiere Yes. – trang1618 Sep 25 '16 at 19:14

These problems can be done easily using l'hospital's rule. Since both the top and the bottom of the limit go to zero, we take the derivative in $h$ to see the limit is:

$lim_{h \rightarrow 0} \frac{f'(x+h) - f'(x-h)}{2h}$

Applying it again:

$= lim_{h \rightarrow 0} \frac{f''(x+h) + f''(x-h)}{2} = f''(x)$

• In order to obtain the last equality, you need to assume not only that $\;f''\;$ exists at the desired point, but also that it is continuous there... – DonAntonio Sep 25 '16 at 19:25
• The idea is fine, but better just do Hôpital once, and then from your first limit deduce directly that $f'$ differentiable at $x$ implies that the limit is $f''(x)$ (e.g. by splitting it into the sum of two limits both converging to $f''(x)/2$ – H. H. Rugh Sep 25 '16 at 19:30
• @DonAntonio Good point. Thanks for pointing that out. – 3-in-441 Sep 25 '16 at 19:33

It's not true in general! It is true in the common situation that $f$ is twice differentiable at $x$, but the limit can exist even when $f''(x)$ does not.

Here are two pathological examples.

The limit can exist even when $f$ is not continuous at $x$. Let $f$ be

$$f(x) = \begin{cases} 0 & x\text{ is rational} \\ 1 & x > 1 \text{ and x is irrational} \\ -1 & x < 1 \text{ and x is irrational} \end{cases}$$

then

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

Even if $f$ is assumed to be differentiable, the existence of the limit isn't enough to conclude the second derivative exists. An example is $f(x) = x|x|$, with $f'(x) = 2|x|$. We have

$$\lim_{h \to 0} \frac{f(x+h) + f(x-h) - 2f(x)}{h^2} = \lim_{h \to 0} \frac{h|h| + (-h)|-h| - 2 \cdot 0}{h^2} = \lim_{h \to 0} 0 = 0$$

but $f''(0)$ does not exist.

• You have pinpointed an important lacking in the formulation of the question. Nevertheless, I wouldn't say, as you do right at the beginning, "it is not true in general" because the OP might be induced to transform it into "it"s very rarely true", although most functions that are commonly used are classicaly used are $C^{\infty}$... – Jean Marie Sep 25 '16 at 19:13

As pointed out by user Hurkyl in his answer the formula you have written in your question does not hold in general but it holds when $f''(x)$ exists. Thus a slight modification of the question is needed.

If $f''(x)$ exists then prove that $$f''(x) = \lim_{h \to 0}\frac{f(x + h) + f(x - h) - 2f(x)}{h^{2}}$$

Any answer which tries to use the definition of $f'$ and $f''$ will lead to iterated limits and hence will not be sufficient to solve the question. The right approach is to use either Taylor's series (as done by user Olivier Moschetta) or via L'Hospital's Rule.

We have \begin{align} L &= \lim_{h \to 0}\frac{f(x + h) + f(x - h) - 2f(x)}{h^{2}}\notag\\ &= \lim_{h \to 0}\frac{f'(x + h) - f'(x - h)}{2h}\text{ (via L'Hospital's Rule)}\tag{1}\\ &= \lim_{h \to 0}\frac{f'(x + h) -f'(x)}{2h} + \frac{f'(x) - f'(x - h)}{2h}\notag\\ &= \frac{f''(x)}{2} + \frac{f''(x)}{2}\tag{2}\\ &= f''(x)\notag \end{align} One should note that the conditions for applicability of L'Hospital Rule hold and therefore we reach step marked $(1)$ by differentiating both denominator and numerator with respect to $h$ (the variable which is used for limit operation and not $x$). It is not possible to apply L'Hospital's Rule again after step $(1)$ as the conditions of its applicability do not hold (we don't know if the second derivatives of $f$ exists at points like $x - h$ or $x + h$, we only know that $f''(x)$ exists). Hence we make use of the definition of $f''(x)$ and accordingly add and subtract $f'(x)$ in the numerator and thereby reach step marked $(2)$.