Under what general conditions can I show that $\displaystyle \lim_{x \to a^-} f'_-(x)=f'_-(a)$? I have derived the following theorems :

*

*If $\displaystyle \lim_{x \to a} f'(x)$ exists and $f$ is continuous at $a$, then $\displaystyle \lim_{x \to a} f'(x)=f'(a)$


*If $\displaystyle \lim_{x \to a^+} f'(x)$ exists and $f$ is continuous at $a$, then $\displaystyle \lim_{x \to a^+} f'(x)=f'(a)$


*If $\displaystyle \lim_{x \to a^-} f'(x)$ exists and $f$ is continuous at $a$, then $\displaystyle \lim_{x \to a^-} f'(x)=f'(a)$
However, I recently saw an argument that had a step using the following claim:

If $\displaystyle \lim_{x \to a^-} f_-'(x)$ exists and $f$ is continuous in some interval around $a$, then $\displaystyle \lim_{x \to a^-} f'_-(x)=f'_-(a)$, where $f'_-$ denotes the left derivative $(*)$.

I looked back at my proofs for 1,2, and 3 and saw that the theorems I used in these proofs (e.g. the mean value theorem) are not applicable for $(*)$. Under what general circumstances is $(*)$ valid, and how does the proof proceed?
 A: Let $f:[a,b] \to \Bbb R$ be a continuous function. An immediate consequence of the mean-value theorem is that if $f$ is differentiable on $(a, b)$ with non-negative (non-positive) derivative then $f$ is increasing (decreasing). Here and in the following, increasing/decreasing is meant in the weak sense.
This can be generalized to functions having left or right derivatives in $(a, b)$ only:

Let $f:[a,b] \to \Bbb R$ be a continuous function.
If $f$  has a right derivative at every point in $(a, b)$ then
$$ \tag{$1$}
 \left ( \forall x \in (a, b): f_+'(x) \ge 0 \right) \implies \text{$f$ is increasing on $[a, b]$.}
$$
$$ \tag{$2$}
 \left ( \forall x \in (a, b): f_+'(x) \le 0 \right) \implies \text{$f$ is decreasing on $[a, b]$.} 
$$
If $f$  has a left derivative at every point in $(a, b)$ then
$$\tag{$3$}
 \left ( \forall x \in (a, b): f_-'(x) \ge 0 \right) \implies \text{$f$ is increasing on $[a, b]$.} 
$$
$$ \tag{$4$}
 \left ( \forall x \in (a, b): f_-'(x) \le 0 \right) \implies \text{$f$ is decreasing on $[a, b]$.} 
$$

For a proof of $(1)$ see for example $f : (0,1) \rightarrow \mathbb{R}$ countinous with non-negative right-hand derivative is non-decreasing, the other variants follow easily.
Using the above we can prove the following:

If $f:[a, b] \to \Bbb R$ is continuous and has a left derivative at each point in $(a, b)$ then
$$ \tag{$5$}
f_+'(a) = \lim_{x \to a^+} f_-'(x) 
$$
$$ \tag{$6$}
f_-'(b) = \lim_{x \to b^-} f_-'(x)
$$
if the limit on the right exists. Similarly, if $f$ has a right derivative at each point in $(a, b)$ then
$$ \tag{$7$}
f_+'(a) = \lim_{x \to a^+} f_+'(x) 
$$
$$ \tag{$8$}
f_-'(b) = \lim_{x \to b^-} f_+'(x) 
$$
if the limit on the right exists.

$(6)$ is what you are asking for, so we will prove that. The other variants can be proved in the same way or can be reduced to $(6)$.
Let us assume that the limit
$$
 \lim_{x \to b^-} f_-'(x) = A
$$
exists. Given $\epsilon > 0$ there is a $x_0 \in (a, b)$ such that
$$ \tag{$*$}
 A - \epsilon < f_-'(x) < A + \epsilon
$$
for $x_0 < x < b$. So the left derivative of $f(x) - (A+\epsilon)x$ is $\le 0$ on $(x_0, b)$. Using $(4)$ above, one can conclude that $f(x) - (A+\epsilon)x$ is decreasing on $[x_0, b]$, so that
$$
 f(x) - (A+\epsilon)x \ge f(b) - (A+\epsilon)b
\implies \frac{f(x)-f(b)}{x-b} \le A + \epsilon \, .
$$
In the same way one uses the left inequality in $(*)$ and $(3)$ to show that
$$
 \frac{f(x)-f(b)}{x-b} \ge A - \epsilon 
$$
for $x_0 < x < b$.
So we have shown that for every $\epsilon > 0$ there is a $x_0 \in (a, b)$ such that
$$
 A - \epsilon \le \frac{f(x)-f(b)}{x-b} \le A + \epsilon 
$$
and that proves that $f$ has a left derivative at $x=b$ with
$$
 f_-'(b) = A = \lim_{x \to b^-} f_-'(x) \, .
$$

Remark: As Paramanand Singh pointed out, requiring continuity at the endpoints of the interval is important. Otherwise counterexamples for all of the above can be found easily, e.g. by defining $f(a)= f(b) = 1$, $f(x) = 0$ otherwise.
The underlying reason is that the mean-value theorem, Rolle's theorem and the above generalizations $(1)-(4)$ all go back to the fact that a continuous function on a compact interval attains its maximum and its minimum. And if the left (right) derivative exists at a maximum then it must be non-negative (non-positive). At a minimum it is the other way around.
