If $f:(a,b)\to \mathbb{R}$ is differentiable in a point $x\in (a,b)$, show that $\lim\limits_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}$ exists

Question

If $f:(a,b)\to \mathbb{R}$ is differentiable in a point $x\in (a,b)$, show that $\lim\limits_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}$ exists and that it is equal to $f'(x)$.

I figured that since the change in $x$ is described trough $2h$ that there is a $\delta$-neighbourhood around $x$ such that $|x-h|=|x+h|$. Sadly, I don't know how to make use of that argument in my proof. I know for a fact that the quotient describes to neighborhoods that, when $h\to 0$, squeeze the point.

Also: https://math.stackexchange.com/q/1680671/42969, https://math.stackexchange.com/q/2394969/42969 – all found with Approach0 — Martin R, Sep 24 '19 at 12:22

score 2 · Accepted Answer · answered Sep 24 '19 at 12:19

2

Hint:

Just consider $$\frac{f(x+h)-f(x-h)}{2h} =\frac{1}{2}\left(\frac{f(x+h)-f(x)}{h} + \frac{f(x)-f(x-h)}{h}\right) $$

answered Sep 24 '19 at 12:19

trancelocation

32,243

Oh god, I should have thought about the "Where does it come from" rather than "what does it describe"... thanks! – Analysis Sep 24 '19 at 12:21
you are welcome :-) – trancelocation Sep 24 '19 at 12:23

If $f:(a,b)\to \mathbb{R}$ is differentiable in a point $x\in (a,b)$, show that $\lim\limits_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}$ exists

1 Answers1