Prove that if $\lim_{x\to a^+}f'(x)$ exists then $\lim_{x\to a^+}f(x)$ exists. Prove that if $\lim_{x\to a^+}f'(x)$ exists then $\lim_{x\to a^+}f(x)$ exists.
I am a high school math teacher, not a professional mathematician. I just thought of this question, in the context of trying to show that if $f''(a)>0$ then the curve does not necessarily have a U-shape around $x=a$, but that is not important here and I am not inviting discussion of that; I just mentioned that to add context.
I have searched for this question, here at MSE and also approachzero, but haven't found anything helpful.
The statement I'm trying to prove here seems intuitively obvious to me but difficult to prove. I have tried using the formal definition of limit, and differentiation from first principles, to no avail.
(I just asked a related question which turned out to be a flawed question, because what I was asking to be proved, is not true. Responders there said that the case with one-sided limits is true, so assuming that is correct, I am asking about that here. In that related question, I showed my flawed attempt at the proof.)
Feel free to let me know if there's anything I can do to improve this question.
 A: Let $(a_n)$ be a sequence converging to $a$ with $a_n >a$, say $a_n=a+\frac 1 n$. Then, by MVT, $f(a_n)-f(a_m)=f'(t)(a_n-a_m)$ for some $t$ between $a_n$ an $a_m$. Let $l =\lim_{x \to a} f'(x) $. There exists $\delta >0$ such that  $|f'(x)|<l+1$ if $x>a,|x-a| <\delta$. If $n$ and $m$ are sufficiently large then $|t-a| <\delta$ so $|f(a_n)-f(a_m)| <(l+1)|a_n-a_m| \to 0$. Hence, $f(a_n)$ is Cauchy sequence. Let $L=\lim f(a_n)$. If $(x_n)$ is any sequence converging to $a$ from the right we can apply MVT again to show that $f(x_n)-f(a_n) \to 0$. Hence, $f(x_n) \to L$. It follows that $f(x) \to L$ as $ x \to a$.
A: Since Kavi already demonstrated a simple proof, let me briefly explain why we expect the result to hold:

*

*Suppose that $f : (a, b) \to \mathbb{R}$ is differentiable and $f'(x)$ converges as $x \to a^+$. Then $f'(x)$ is bounded near $a$, that is, there exist $c \in (a, b)$ and $M \in (0, \infty)$ such that $|f'(x)| \leq M$ whenever $x \in (a, c)$.


*A differentiable function on an interval with bounded derivative is an example of uniformly continuous function. Recall that a function $f : E \subseteq \mathbb{R} \to \mathbb{R}$ is said to be uniformly continuous if
$$ \forall \varepsilon > 0, \ \exists \delta > 0, \ \text{s.t.} \ \forall x, y \in E \ : \, |x - y| < \delta \implies |f(x) - f(y)| < \varepsilon. $$
(That is, uniformly continuity requires that a single choice of $\delta$ should work for any point.) Then it is well-known that:

Theorem. Any uniformly continuous $f : E \subseteq \mathbb{R} \to \mathbb{R}$ extends to a uniformly continuous function on $\overline{E}$, the closure of $E$.

These two observations together immediately prove OP's claim.
