I don't understand in the proof below why you can claim that the derivative the function in $x_0$ of $h\to 0$ for $h > 0$, is smaller than or equal to zero. From what rule this follows? Does it follow from calculating the limit? How?
From wikipedia: https://en.wikipedia.org/wiki/Fermat%27s_theorem_(stationary_points)
Proof 2: Extremum implies derivative vanishes
Alternatively, one can start by assuming that $\displaystyle x_0$ is a local maximum, and then prove that the derivative is 0.
Suppose that $\displaystyle x_0$ is a local maximum (a similar proof applies if $\displaystyle x_0$ is a local minimum). Then there $\exists \, \delta > 0 $ such that $(x_0 - \delta,x_0 + \delta) \subset (a,b)$ and such that we have $f(x_0) \ge f(x)\, \forall x$ with $\displaystyle |x - x_0| < \delta $. Hence for any $h \in (0,\delta)$ we notice that it holds
:$\frac{f(x_0+h) - f(x_0)}{h} \le 0.$
Since the [[Limit of a function|limit]] of this ratio as $\displaystyle h$ gets close to 0 from above exists and is equal to $\displaystyle f'(x_0)$ we conclude that $f'(x_0) \le 0$. On the other hand for $h \in (-\delta,0)$ we notice that
:$\frac{f(x_0+h) - f(x_0)}{h} \ge 0$
but again the limit as $\displaystyle h$ gets close to 0 from below exists and is equal to $\displaystyle f'(x_0)$ so we also have $f'(x_0) \ge 0$.
Hence we conclude that $\displaystyle f'(x_0) = 0.$
When $h\in(0,\delta)$, we know that $f(x_0+h)\leq f(x_0)$, since $x_0$ is a maximum in $(x_0-\delta,x_0+\delta)$. So $f(x_0+h)-f(x_0)\leq 0$. Since $h$ is positive, we get that:
$$\frac{f(x_0+h)-f(x_0)}{h}\leq 0$$
So $$f'(x_0)=\lim_{h\to 0^+} \frac{f(x_0+h)-f(x_0)}{h}\leq 0$$
When $h\in(-\delta,0)$ then, again we have have: $f(x_0+h)\leq f(x_0)$ and thus $f(x_0+h)-f(x_0)\leq 0$. But now, when we divide by negative $h$, we have to reverse the signs, so we get:
$$\frac{f(x_0+h)-f(x_0)}{h}\geq 0$$
So you get:
$$f'(x_0)=\lim_{h\to 0^-} \frac{f(x_0+h)-f(x_0)}{h}\geq 0$$