The following is an exercise from Calculus by Spivak.
Suppose that $f:\mathbb{R}\to\mathbb{R}$ is differentiable at some point $x\in\mathbb{R}$. Prove that $$f'(x)=\lim_{h,k\to 0^+}\frac{f(x+h)-f(x-k)}{h+k}$$
My attempt - I tried something with substitutions, but it doesn't feel rigorous enough:
Let $a=x+h$, and $b=x-k$. Then we have \begin{align*} \lim_{h,k\to 0^+}\frac{f(x+h)-f(x-k)}{h+k} & =\lim_{k\to 0^+}\Bigg[\lim_{h\to 0^+}\frac{f(x+h)-f(x-k)}{h+k}\Bigg] \\ & = \lim_{b\to x^-}\Bigg[\lim_{a\to b}\frac{f(a)-f(b)}{a-b}\Bigg] \\ & = \lim_{b\to x^-}f'(b) \\ & = f'(x) \end{align*}
This just feels quite off, but I wasn't sure how else to approach it. Any help is appreciated.
$$ \begin{split} \frac{f(x+h)-f(x-k)}{h+k} &= \frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k} \\ &= \frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} \xrightarrow[\substack{h\to0^+\\h\to0^+}]{} f'(x) . \end{split} $$