TITLE: Is this function concave?
QUESTION [1 upvotes]: I was trying to solve an excersise when I bumped into this:

If a function $f$ : $\mathbb{R} \to\mathbb{R}$ is strictly increasing and differentiable and it is true that 
  $\lim_{x\to\infty} f(x)=l$ where $l\in\mathbb{R}$, then $f$ is concave at some interval of the form $(k,+\infty)$. 

It seems so obvious to me yet I have been struggling way to much to prove it. Any help would be appreciated.

REPLY [1 votes]: Counterexample: Define
$$ f(x) = \int_0^x \frac{2+\sin t}{1+t^2}\, dt.$$
Then $f\in C^\infty(\mathbb R).$ Note that the above integral converges if $x=\infty,$ showing that $\lim_{x\to \infty} f(x)$ exists. By the FTC, $f'(x) = (2+\sin x)/(1+x^2) > 0,$ so $f$ is strictly increasing. Differentiating again, we get
$$\tag 1 f''(x) = \frac{(1+x^2)\cos x- (2+\sin x)(2x)}{(1+x^2)^2}.$$
Along the sequence $2n\pi,$ the numerator in $(1)\to \infty,$ so $f''(2n\pi) > 0$ for large $n.$
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