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I read somewhere recently that you can define the derivative as follows:

$$f'(x) = \lim_{h, k \to 0^+} \frac{f(x +h) - f(x - k)}{h + k}$$

I have been trying to prove this for about 2 hours, and can't seem to get it done. How should I proceed?

Edit: Assume $f$ is differentiable for all $x \in (x - \delta, x + \delta)$ for some $\delta > 0$.

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1 Answer 1

up vote 9 down vote accepted

It is not hard to show that if the limit given in the post exists, then the derivative exists, and is equal to the given limit. Just let $k=h$, and rewrite the expression as $$\frac{1}{2}\frac{f(x+h)-f(x)}{h}+\frac{1}{2}\frac{f(x-h)-f(x)}{-h} .$$

For the converse, suppose $f'(x)$ exists. Rewrite our expression as $$\frac{h}{h+k}\frac{f(x+h)-f(x)}{h} +\frac{k}{h+k}\frac{f(x-k)-f(x)}{-k}.\tag{$1$}$$

Choose an $\epsilon \gt 0$. Because $f'(x)$ exists, there is a $\delta$ such that if $|t|\lt \delta$, then $$\left|\frac{f(x+t)-f(x)}{t}-f'(x)\right|\lt \epsilon.$$

It follows that if $h$ and $k$ are $\lt \delta$, then the differential quotients in $(1)$ are each within $\epsilon$ of $f'(x)$. Thus $(1)$ is equal to $$\frac{h}{h+k}(f'(x)\pm\epsilon_1)+\frac{k}{h+k}(f'(x)\pm\epsilon_2),$$ where $\epsilon_1$ and $\epsilon_2$ are non-negative quantities $\lt \epsilon$. But $h/(h+k)$ and $k/(h+k)$ are both less than $1$. (This little fact is key: we could run into trouble using two points on the same side of $x$.)

It follows that if $h$ and $k$ are less than $\delta$, then $(1)$ differs from $f'(x)$ by less than $\epsilon$. So the limit as $h$, $k$ independently go to $0^+$ exists.

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Sorry, assume that $f$ is differentiable at and in a neighbourhood of $x$. I will put this in an edit above. –  providence Nov 24 '12 at 1:15
    
Wow, I am punching myself in the head. –  providence Nov 24 '12 at 1:19
    
@providence: I had an error, forgot the points are on opposite sides of $x$. Then $x^2\sin(1/x)$ is a problem. But with $h$, $k$ both positive, everything is fine. –  André Nicolas Nov 24 '12 at 2:32

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