If $f$ differentiable at $a$, $\lim\limits_{x \to a^+} \frac{f(x)-f(a)}{x-a} \geq 0$, what is the exact argument that allows us to say $f'(a) \geq 0$? The following is a problem from Ch. 11 in Spivak's Calculus:



*(a) Suppose that $f$ is differentiable on $[a,b]$. Prove that if the minimum of $f$ on $[a,b]$ is at $a$, than $f'(a) \geq 0$, and if
it is at $b$, than $f'(b) \leq 0$

Proof
Let $x \in (a,b)$.
Then since $a$ is a minimum of $f$ on $[a,b]$, we have that $\frac{f(x)-f(a)}{x-a} \geq 0$.
Therefore,
$$\lim\limits_{x \to a^+} \frac{f(x)-f(a)}{x-a} \geq 0$$
$f$ is differentiable at $a$ by assumption. Therefore,
$$\lim\limits_{x \to a^-} \frac{f(x)-f(a)}{x-a}=\lim\limits_{x \to a^+} \frac{f(x)-f(a)}{x-a} \geq 0$$
At this point we need to conclude that $f'(a) \geq 0$. My question is: what is the exact argument that allows us to conclude this?
Is the argument simply that because $f$ is differentiable at $a$, we know that the limits defining the derivative from above and below at $a$ must be equal. So since the limit from above is $\geq 0$, the limit from below must be as well, and therefore $f'(a) \geq 0$ as well.
In graph to the right below, with the extra information that $a$ is a minimum also on an interval $(a-\delta, a]$, we could conclude that
$$\lim\limits_{x \to a^-} \frac{f(x)-f(a)}{x-a} \leq 0$$
We can see from the graph that in fact $f'(a)=0$.
We would have
$$\lim\limits_{x \to a^+} \frac{f(x)-f(a)}{x-a} \geq 0$$
$$\lim\limits_{x \to a^-} \frac{f(x)-f(a)}{x-a} \leq 0$$
So we would be able to say that $0 \leq f'(a) \leq 0$

 A: 
Proof
Let $x\in(a,b)$.
Then since $a$ is a minimum of $f$ on $[a,b]$, we have that $\dfrac{f(x)−f(a)}{x−a}\geq0$.
Therefore,
$$\lim_{x\to a^{+}}\frac{f(x)−f(a)}{x−a}\geq0$$

And you're done!
Modulo some minor inaccuracies in wording (for example, I'd rather say that "since the minimum of $f$ on $[a,b]$ is at $a$", as it was in the statement of the problem), this is all you needed to do.
You can't set up "$\displaystyle \lim_{x\to a^{-}}\frac{f(x)−f(a)}{x−a}$" here, as the function $f$ is only defined on $[a,b]$, and therefore $f(x)$ for $x$ to the left of $a$ doesn't make sense.
When we say that a function is differentiable on a closed, bounded interval $[a,b]$, by definition it means that the usual derivative exists for all interior points $x\in(a,b)$, but the derivative at $a$ is understood as the right-sided derivative and the derivative at $b$ is understood as the left-sided derivative.
A: Let's rewrite the proof as follows:

Let $x \in (a,b)$.
Then since $a$ is a minimum of $f$ on $[a,b]$, we have that $\frac{f(x)-f(a)}{x-a} \geq 0$.
Therefore,
$$\lim_{x \to a^+} \frac{f(x)-f(a)}{x-a} \geq 0$$
$f$ is differentiable at $a$ by assumption. Therefore,
$$f'(a) = \lim_{x\to a}\frac{f(x)-f(a)}{x-a} = \lim_{x \to a^+} \frac{f(x)-f(a)}{x-a} \geq 0$$


The only step we changed was the end.  Specifically, this just uses the fact that if $\lim_{x\to c} g(x)$ exists, then $\lim_{x\to c} g(x) = \lim_{x\to c^+} g(x)$
A: It looks like you are not using that the function is defined on $[a,b]$. So, the lower limit is not defined, and the derivative at $a$ is just given by the upper limit. So $f’(a)\geq0$ by what you have already shown, and it does not have to equal $0$.
