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Assume $A$ is open. If $g$ is differentiable at $c\in A$, show that $g'(c)=\lim_{h\to0}\frac{g(c+h)-g(c-h)}{2h}$.

With a little algebraic manipulation we can see that

$$\frac{g(c+h)-g(c-h)}{2h}=\frac{1}{2}\bigg(\frac{g(c+h)-g(c)}{h} + \frac{g(c)-g(c-h)}{h}\bigg) \stackrel{?}{=} \frac{1}{2}\bigg(2g'(c)\bigg)$$

The first summand is the difference quotient, so by definition it's equal to $g'(c)$ when $h\to0$. The second summand is a disguised different quotient, but we could raise questions regarding the signs of these summands (e.g. could they cancel each other out?).

However, if we study the lateral limits, we see that

$$\lim_{h\to 0^-} \frac{g(c+h)-g(c)}{h} = \lim_{h\to 0^+} \frac{g(c)-g(c-h)}{h} \\ \lim_{h\to 0^+} \frac{g(c+h)-g(c)}{h} = \lim_{h\to 0^-} \frac{g(c)-g(c-h)}{h}$$

Since the left-hand side of this pair of identities is $g'(c)$, the right-hand side should also be identical to each other and to $g'(c)$. Is this argument sound?

Also, why do we need to suppose that $A$ is open? What would change if $A$ were closed?

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    $\begingroup$ The openness of $A$ is needed to ensure the limit makes sense. If $A$ is open, we can find $h_0$ small enough such that $c+h\in A$ for all $|h|<h_0$, so $g(c+h)$ and then the limit make sense. Here I assume “$a\in A$” in your problem is to be “$c\in A$”. $\endgroup$ – Feng Shao Jul 20 at 1:18
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We want to prove that $$\lim_{h\to 0} \frac{g(c)-g(c-h)}{h}=g'(c)$$ If we take $k:= -h$, then $k\to 0$ as $h\to 0$, and $$\lim_{h\to 0} \frac{g(c)-g(c-h)}{h}=\lim_{k\to 0} \frac{g(c)-g(c+k)}{-k}=\lim_{k\to 0} \frac{g(c+k)-g(c)}{k}=g'(c)$$ More general, first prove with the $\varepsilon,\delta$ definition that $$\lim_{x\to 0}f(x)=\lim_{x\to 0}f(-x)$$

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