As note din the comments by many, there are two things at issue here.
First, you cannot take a complex expression and only evaluate the limit to part of it, and then to the rest of it. If you have something like
$$\lim_{x\to 0}\frac{x}{x},$$
the limit is $1$ (since the function takes the value $1$ at every $x\neq 0$); but you can't first evaluate the limit of the numerator (which is $0$), and then the limit of the resulting expression to claim that the limit is $0$.
So you cannot first do the limit of the numerator of
$$\frac{F(x+h)-F(x)}{h}$$
and then do the expression. You need to do the whole expression. That is: you cannot compute the limit piecemeal within an expression: you must compute the limit of the whole, or of all its parts, at the same time.
Second, there is a "limit law" (actually a theorem for evaluating limits) that says:
Theorem. Let $g(x)$ and $h(x)$ be function, and assume that $\lim\limits_{x\to a}g(x)=L$ and $\lim\limits_{x\to a}h(x) = M$ both exist. Then:
- $\lim\limits_{x\to a}\bigl(g(x)+h(x)\bigr) = L+M$;
- $\lim\limits_{x\to a}g(x)h(x) = LM$;
- If $M\neq 0$, then $\lim\limits_{x\to a}\frac{g(x)}{h(x)} = \frac{L}{M}$.
This can be established fairly straightforwardly using the $\epsilon$-$\delta$ definition of a limit, though there is a bit of technical work to be done in 3 to make sure that $h(x)$ is bounded away from $0$. These limit laws are sometimes shorthanded as
The limit of a sum is the sum of the limits (if they both exist); the limit of the product is the product of the limits (if they both exist); the limit of a quotient is the quotient of the limits if the limit of the denominator is not $0$ (and both exist).
We can use these "limit laws" in the right hand side: the denominator is constant and has limit $2\neq 0$; so
$$\lim{h\to 0}\frac{f(x+h)+f(x)}{2} = \frac{\lim\limits{h\to 0}\bigl( f(x+h)+f(x)\bigr)}{\lim_{h\to 0}2} = \frac{f(x)+f(x)}{2} = f(x),$$
using the limit law and the fact that $f(x)$ is assumed to be continuous everywhere in the interval, so $\lim\limits_{h\to 0}f(x+h) = f(x)$; in fact, we used limit law 1 to compute the limit of the numerator.
You cannot invoke this law/process in the left hand side, because there the denominator has limit $\lim\limits_{h\to 0}h = 0$. So you can't take the limits of the numerator and denominator separately. You have to deal with the entire expression in some way (obviously there are ways of doing it, or no derivative could ever be calculated; but you can't use the limit laws used on the right hand side).
(As an aside, note that while $\lim\limits_{h\to 0}f(x+h) = f(x)$ because we are assuming that $f$ is continuous, we have not proven that $F$ is continuous, so you also can't really justify the claim that $\lim\limits_{h\to 0}F(x+h)=F(x)$ unless you prove the continuity.)
Best Answer
I will expand on my comment slightly. The key to this is monotonicity. To avoid boundary cases, I'll consider $x_0 \in (a,b)$. Suppose $f$ is discontinuous at $x_0$. Let $L^- = \lim_{x \to x_0^-} f(x)$ and $L^+ = \lim_{x \to x_0^+} f(x)$. Since $f$ is monotonically increasing, we have that $$f(y) \le L^- \le L^+ \le f(x)$$ for all $y < x_0 < x$. Moreover, discontinuity ensures that $L^- < L^+$.
I would suggest that you take $\alpha,\beta \in (L^-,L^+)$, $\alpha < \beta$, and estimate the values $$\int_{x_0}^{x_0 + \epsilon} f \qquad \text{and} \qquad \int_{x_0-\epsilon}^{x_0} f $$ for $\epsilon > 0$ sufficiently small. Can you show that the one-sided limits $$\lim_{\epsilon \to 0^+} \frac{F(x_0-\epsilon) - F(x_0)}{-\epsilon} \qquad \text{and} \qquad \lim_{\epsilon \to 0^+} \frac{F(x_0+\epsilon) - F(x_0)}{\epsilon}$$ are different?