Proving limit of f(x), f'(x) and f"(x) as x approaches infinity

[WWGD]

Maybe this could move things forward:
Try the mvt with a Telescope:
Find the least integer a' larger than a. Then use
$\Sigma f(a+1)-f(a)=\Sigma f'(a_*) ; a_*\in (a, a+1)$ , etc.( Since $f$ is assumed differentliable in $( a, \infty))$That would show , by properties of Telescopes, since f' is finite, thst the nth term goes to 0. A similar idea for

I'd appreciate a basis for your doubt Is it not the case that

$Lim _{x \rightarrow \infty} \Sigma_{j=1}^{\infty} (a_{j+1} -a_j) = a_1 - lim_{ j \rightarrow \infty} a_j$,
which in our case equals $f(1)- Lim_{ n \rightarrow \infty} f'(x_n)?$, which converges iff the sequence $f_n:= f'(x_n) ; x_n:=(a +n)$ converges, or do you oppose it on a different basis?

 

[Office_Shredder]

@WWGD that only constructs a single unbounded sequence of points for which the derivative converges to zero. How do you handle the rest of them?

 

[WWGD]

@WWGD that only constructs a single unbounded sequence of points for which the derivative converges to zero. How do you handle the rest of them?

I agree; it was just an idea for a path ahead to show $Lim_{x \infty}f'(x)=0$. I'm working on the details, using sequential continuity , given $f \in C^2(a, \infty)$. But maybe I should have been explicit that it was not intented as a worked solution.