Spivak Calculus, Chapter 9, Exercise 22(b) Assuming $f : \mathbb R \to \mathbb R$ is differentiable, how can one prove that
$$
f'(x) = \lim_{h,k \to 0^+} \frac{f(x+h)-f(x-k)}{h+k},
$$
an alternate expression to the usual limit definition of the derivative $f'(x)=\lim_{h \to 0^+} \frac{f(x+h)-f(x)}h$?
I figured the problem out for the special case of $k=h$: add and subtract $f(x)$ in the numerator of the expression. An $\epsilon$-$\delta$ requirement was not required to prove that $\frac{f(x+h)-f(x-h)}{2h} = \frac{f(x+h)-f(x)}h$. This was part (a) of Exercise 9.22.
I also thought about sending $k \to 0^+$ first to simplify the function, and since $f$ is differentiable (therefore continuous), sending $k \to 0^+$ would reduce the original expression down to the familiar $\lim_{h \to 0^+} \frac{f(x+h)-f(x)}{h}$. But then again we can't send $h \to 0^+$ and $k \to 0^+$ separately.
 A: HINT: You were on the right track to subtract and add $f(x)$ in the numerator. So then you'll have the sum of two fractions. You now need to do a similar trick with multiplication with each of those factors in order to make the usual difference quotients
$$\frac{f(x+h)-f(x)}h \qquad\text{and}\qquad \frac{f(x+k)-f(x)}k$$
appear.
A: It's better to use an $\varepsilon$-$\delta$ argument to present a rigorous proof: that is, evaluating the difference between the operand and the objective $f'(x)$.

Details:
Given $\varepsilon > 0$, by the definition of $f'(x)$, there exists $\delta > 0$ such that $0 < h < \delta$ and $0 < k < \delta$ imply that 
$$\left|\frac{f(x + h) - f(x)}{h}  - f'(x)\right| < \varepsilon,\qquad\left|\frac{f(x - k) - f(x)}{-k} - f'(x)\right| < \varepsilon.$$
Now if $\|(h, k) - (0, 0)\| < \delta$, we then have:
\begin{align*}
 & \left|\frac{f(x + h) - f(x - k)}{h + k}  - f'(x)\right|\\
= & \left|\frac{f(x + h) - f(x)}{h} \cdot \frac{h}{h + k} - f'(x)\frac{h}{h + k} + \frac{f(x - k) - f(x)}{-k} \cdot \frac{k}{h + k} - f'(x)\frac{k}{h + k}\right| \\ 
\leq & \left|\frac{f(x + h) - f(x)}{h}  - f'(x)\right| \frac{h}{h + k} + \left|\frac{f(x - k) - f(x)}{-k} - f'(x)\right|\frac{k}{h + k} \\
< & \varepsilon \frac{h}{h + k} + \varepsilon \frac{k}{h + k} \\
= & \varepsilon.
\end{align*}
This is precisely what we want to prove.
Note the condition $h > 0$ and $k > 0$ is crucial whence we can remove the absolute value.
A: This is too long for a comment, so I'm posting here my understanding of answering this problem:  
Since by hypothesis $f'(x)=\lim_{h \to 0^+} \frac{f(x+h)-f(x)}h = \lim_{k \to 0^+} \frac{f(x)-f(x-k)}k$...
For all $\epsilon > 0$, there exists $\delta_1 > 0$ such that $0 < h-0<\delta_1$ implies $\left|\frac{f(x+h)-f(x)}h-f'(x)\right|<\epsilon$. 
For all $\epsilon > 0$, there exists $\delta_2 > 0$ such that $0 < k-0 <\delta_2$ implies $\left|\frac{f(x)-f(x-k)}k-f'(x)\right|<\epsilon$. 
Choose  $\require{enclose}
     \enclose{horizontalstrike}{\delta := \min\{\delta_1,\delta_2\}}$. Choose $\delta_1=\delta_2=\epsilon$. Then
\begin{align}
&\left|\frac{f(x+h)-f(x-k)}{h+k}-f'(x) \right| \\ 
&\le \frac{h}{h + k} \left|\frac{f(x + h) - f(x)}{h}  - f'(x)\right| + \frac{k}{h + k} \left|\frac{f(x) - f(x-k)}{k} - f'(x)\right| \\ 
&= \frac{h}{h + k} \left|\frac{f(x + h) - f(x)}{h}  - f'(x)\right| + \frac{k}{h + k} \left|\frac{f(x+k) - f(x)}{k} - f'(x)\right| \\ 
&< \frac{h}{h+k} \delta_1 + \frac{k}{h+k} \delta_2 \\ 
&= \frac{h}{h+k} \epsilon + \frac{k}{h+k} \epsilon \\ 
&= \epsilon
\end{align}
