If f'(a) exists, show that $f'(a) = \lim_{h\to 0} \frac {f(a+h) - f(a-h)}{2h}$ I have proven it in terms of a, by using $1/2 \lim_{h\to 0} \frac {f(a+h)-f(a)}{h} + 1/2 \lim_{h\to 0} \frac {f(a-h)-f(a)}{-h} = f'(a)$.
How do you prove this in terms of $\varepsilon$ and $\delta$.
On
You can just sum and subtract $f(a)$ in the numerator and solve the limit directly. But I'll indulge you: let $\epsilon > 0$. Since you know that the derivative exists, there is $\delta>0$ such that $$0<|h|<\delta\implies\left|\frac{f(a+h)-f(a)}{h}-f'(a)\right|<\epsilon. $$Now, assume $0 <|h|<\delta$. Then $$\begin{align} \bigg|\frac{f(a+h)-f(a-h)}{2h}&- f'(a)\bigg| = \left|\frac{f(a+h)-f(a)+f(a)-f(a-h)}{2h} - f'(a)\right| \\ &= \left|\frac{f(a+h)-f(a)}{2h} - \frac{f'(a)}{2} +\frac{f(a)-f(a-h)}{2h} - \frac{f'(a)}{2}\right| \\ &\leq \frac{1}{2} \left|\frac{f(a+h)-f(a)}{h}-f'(a)\right|+\frac{1}{2}\left|\frac{f (a)-f (a-h)}{h}-f'(a)\right| \\ &=\frac{\epsilon}{2} +\frac{\epsilon}{2} = \epsilon,\end{align} $$as wanted.
Given $\epsilon>0$, we have some $\delta>0$ such that $\left|\dfrac{f(a+u)-f(a)}{u}-f'(a)\right|<\epsilon$ for all $0<|u|<\delta$.
For $0<|h|<\delta$, then $0<|-h|<\delta$ of course, so \begin{align*} \left|\dfrac{f(a-h)-f(a)}{-h}-f'(a)\right|<\epsilon, \end{align*} and \begin{align*} \left|\dfrac{f(a+h)-f(a)}{h}-f'(a)\right|<\epsilon, \end{align*} so \begin{align*} &\left|\dfrac{f(a+h)-f(a-h)}{2h}-f'(a)\right|\\ &=\dfrac{1}{2}\left|\dfrac{f(a+h)-f(a)}{h}-f'(a)+\left(\dfrac{f(a-h)-f(a)}{-h}-f'(a)\right)\right|\\ &\leq\dfrac{1}{2}\left[\left|\dfrac{f(a+h)-f(a)}{h}-f'(a)\right|+\left|\dfrac{f(a-h)-f(a)}{-h}-f'(a)\right|\right]\\ &<\dfrac{1}{2}(\epsilon+\epsilon)\\ &=\epsilon. \end{align*}