Solutions

a) Given $f'(x)$ is defined, for any $\epsilon>0$ there exists $\delta>0$ such that

$$\forall h,\,0<|h|<\delta\implies \left|\frac{f(x+h)-f(x)}{h}\right|<\epsilon.$$

We can substitute $h=-k$. This alternatively means that

$$\forall k,\,0<|k|<\delta\implies \left|\frac{f(x-k)-f(x)}{-k}\right|=\left|\frac{f(x)-f(x-k)}{k}\right|<\epsilon$$

and consequently

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

Hence,

$$\begin{align*} \lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h} &=\lim_{h\to 0}\frac{1}{2}\left(\frac{f(x+h)-f(x)}{h}+\frac{f(x)-f(x-h)}{h}\right) \\ &=\frac{1}{2}\left[\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}+\lim_{h\to 0}\frac{f(x)-f(x-h)}{h}\right] \\ &=\frac{1}{2}(f'(x)+f'(x)) \\ &=f'(x). \end{align*}$$

b) By a), we know that

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

We also know that a limit is defined iff its left-hand and right-hand limits are defined. Thus, for any $\epsilon>0$, there exist $\delta_1,\delta_2>0$ such that

$$0<h<\delta_1\implies \left|\frac{f(x+h)-f(x)}{h}-f'(x)\right|<\epsilon$$

and

$$0<k<\delta_2\implies \left|\frac{f(x)-f(x-k)}{k}-f'(x)\right|<\epsilon.$$

For all $h$ and $k$, let $\delta=\min(\delta_1,\delta_2)$. Then, for all $h$ and $k$ such that $0<h<\delta$ and $0<k<\delta$, we have

$$\begin{align*} \left|\frac{f(x+h)-f(x-k)}{h+k}-f'(x)\right| &\le \left|\frac{f(x+h)-f(x)-hf'(x)}{h+k}+\frac{f(x)-f(x-k)-kf'(x)}{h+k}\right| \\ &\le \frac{h}{h+k}\left|\frac{f(x+h)-f(x)}{h}-f'(x)\right|+\frac{k}{h+k}\left|\frac{f(x)-f(x-k)}{k}-f'(x)\right| \\ &<\frac{h}{h+k}\epsilon+\frac{k}{h+k}\epsilon \\ &=\epsilon. \end{align*}$$

Hence,

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