# Proof for alternative definition of the derivative

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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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