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

Question

In the book of Calculus by Spivak, at page 152, question 22, it is asked that

Suppose that $f$ is differentiable at $x$. Prove that

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

I have tried some algebraic trics & take a look at this, question, but couldn't find the exact derivation, so I would appreciate any help or hint.

I mean it is clear that this is a more general case of $$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x-h)}{2h},$$ and we are going to use similar ideas.

Answer 1

If $h,k$ are sufficiently small then

$$f'(x) - \epsilon < \frac{f(x+h)-f(x)}{h} < f'(x) + \epsilon \\ f'(x) - \epsilon < \frac{f(x)-f(x-k)}{k} < f'(x) + \epsilon$$

and

$$ \frac{f(x+h)-f(x-k)}{h+k} = \frac{h}{h+k}\frac{f(x+h)-f(x)}{h}+ \frac{k}{h+k} \frac{f(x)-f(x-k)}{k}$$

The LHS is bounded between $\frac{h+k}{h+k}(f'(x) - \epsilon)$ and $\frac{h+k}{h+k}(f'(x) + \epsilon).$

Hence, if $h,k$ are sufficiently small

$$f'(x) - \epsilon < \frac{f(x+h)-f(x-k)}{h+k} < f'(x) + \epsilon $$

Answer 2

Given $\epsilon>0$, there exists some $\delta>0$ such that for all $0<|u|<\delta$, then $\left|\dfrac{f(x+u)-f(x)}{u}-f'(x)\right|<\epsilon$.

For all $0<h,k<\delta$, then \begin{align*} &\left|\dfrac{f(x+h)-f(x-k)}{h+k}-f'(x)\right|\\ &=\dfrac{1}{h+k}|f(x+h)-f(x)-hf'(x)+[f(x)-f(x-k)-kf'(x)]|\\ &\leq\dfrac{1}{h}|f(x+h)-f(x)-hf'(x)|+\dfrac{1}{k}|f(x)-f(x-k)-kf'(x)|\\ &=\left|\dfrac{f(x+h)-f(x)}{h}-f'(x)\right|+\left|\dfrac{f(x-k)-f(x)}{-k}-f'(x)\right|\\ &<\epsilon+\epsilon\\ &=2\epsilon. \end{align*}

Answer 3

HINT

Note that

• $f(x+h) = f(x)+f’(x)h+\frac12 f’’(x)h^2+o(h^2)$
• $f(x-k) = f(x)-f’(x)k+\frac12 f’’(x)k^2+o(k^2)$