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.
Best Answer
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$.