Let $f:I \longrightarrow \mathbb{R}$ be differentiable in a inner point $\xi \in I$.
Show that $$\lim\limits_{h \rightarrow 0}\frac{f(\xi+h)-f(\xi-h)}{2h}$$
exists and is $f'(\xi)$.
Also give an example which shows, that the existence of this limit, doesnt mean, that $f$ is necessary differentiable in $\xi$.
My attempt:
Given is that:
$$\lim\limits_{h\rightarrow0}\frac{f(\xi+h)-f(\xi)}{h}=f'(\xi)$$
exists.
If $\,\lim\limits_{h \rightarrow 0}\frac{f(\xi+h)-f(\xi-h)}{2h}=\lim\limits_{h\rightarrow0}\frac{f(\xi+h)-f(\xi)}{h}=f'(\xi)$ then:
$$\lim\limits_{h\rightarrow0}\left(\frac{f(\xi+h)-f(\xi)}{h}-\frac{f(\xi+h)-f(\xi-h)}{2h}\right)=0$$
$$\lim\limits_{h\rightarrow0}\frac{2(f(\xi+h)-f(\xi))-f(\xi+h)+f(\xi-h)}{2h}=\lim\limits_{h\rightarrow0}\frac{f(\xi+h)-f(\xi)+f(\xi-h)-f(\xi)}{2h}=0$$
$$=\frac{1}{2}\lim\limits_{h\rightarrow0}\frac{f(\xi+h)-f(\xi)}{h}+\frac{1}{2}\lim\limits_{h\rightarrow0}\frac{f(\xi-h)-f(\xi)}{h}=\frac{1}{2}\lim\limits_{h\rightarrow0}\frac{f(\xi+h)-f(\xi)}{h}-\frac{1}{2}\lim\limits_{h\rightarrow0}\frac{f(\xi)-f(\xi-h)}{h}=\frac{1}{2}f'(\xi)-\frac{1}{2}f'(\xi)=0$$
$\Box$
Let $f:\mathbb{R}\longrightarrow \mathbb{R}:x \mapsto |x|$
Then for $\xi=0$:
$$\lim\limits_{h \rightarrow 0}\frac{f(\xi+h)-f(\xi-h)}{2h}=\lim\limits_{h \rightarrow 0}\frac{|h|-|-h|}{2h}=\lim\limits_{h \rightarrow 0}\frac{|h|-|h|}{2h}=\lim\limits_{h \rightarrow 0}0=0$$
$\Longrightarrow$ the limit exists, but $f$ is not differentiatable at $\xi=0$
Hello, as always it would help me alot, if someone could look over it and give me feedback weither my work is correct. And if not, what is wrong :) thank you
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