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

Question

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

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...

Answer 1

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

Answer 2

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

Answer 3

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)$

Answer 4

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.

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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

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

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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.

Answer 5

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 $(2)$.