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I want to prove that if $f(x)$ has second order derivative than it should given by: $$f''(x)=\lim_{h \to 0} \frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$ but I can not prove this. So I try to write that definition of the derivative:

\begin{equation}\label{eq1} f''(x)= \lim_{h\to0} \frac{f'(x+h)-f'(x)}{h}\quad(1) \end{equation}

From here I expand: $$f'(x+h)= \lim_{c\to0} \frac{f(x+h+c)-f(x+h)}{c}$$ and similarly for $f'(x)$ in $(1)$. So then I substitute it into $(1)$:

\begin{align} f''(x)=\lim_{h\to0} \frac{\lim_{c\to0} \frac{f(x+h+c)-f(x+h)}{c}-\lim_{c\to0} \frac{f(x+c)-f(x)}{c}}{h} \\= \lim_{h\to0} \frac{\lim_{c\to0} [\frac{f(x+h+c)-f(x+h)}{c}- \frac{f(x+c)-f(x)}{c}]}{h} \\= \lim_{h\to0} \frac{\lim_{c\to0} [\frac{f(x+h+c)-f(x+h)- f(x+c)-f(x)}{c}]}{h} \end{align} What I am trying to do is to prove is that $$\lim_{h\to0} \frac{\lim_{c\to0} [\frac{f(x+h+c)-f(x+h)- f(x+c)-f(x)}{c}]}{h} = \lim_{h \to 0} \frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$ but I could not do it.

A P
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Alex Quchuloria
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    https://math.stackexchange.com/q/210264/42969, https://math.stackexchange.com/q/491248/42969 – Martin R Feb 08 '24 at 15:25
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    I would take $c \rightarrow h$, instead of $c \rightarrow 0$. – ck1987pd Feb 08 '24 at 17:11
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    Notice that this is not a limit definition of the second derivative; it's possible for this limit to exist even if the second derivative does not. (Example: $ f ( x ) = x ^ 3 \sin ( 1 / x ) $ with $ f ( 0 ) = 0 $, at $ x = 0 $.) But if $ f ' ' ( x ) $ does exist, then it must equal this limit; that's all that you can expect to prove. – Toby Bartels Feb 08 '24 at 19:44

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Notice that $h$ and $c$ both approach $0$ at the same time, so you could replace both with a variable $k$ approaching $0.$

This means that $$\lim_{h\to0} \frac{\lim_{c\to0} [\frac{f(x+h+c)-f(x+h)- f(x+c)-f(x)}{c}]}{h} = \lim_{k\to0} \frac{f(x+2k)-2f(x+k)-f(x)}{k^2}.$$

Cyclonestopper
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    This kind of thing is dangerous; for example, if I have $ \lim \limits _ { x \to 0 } \lim \limits _ { y \to 0 } \frac { 2 x y } { x ^ 2 + y ^ 2 } $, then this is $ 0 $, but if I change it to $ \lim \limits _ { k \to 0 } \frac { 2 k k } { k ^ 2 + k ^ 2 } $, then this is $ 2 $. – Toby Bartels Feb 08 '24 at 19:33