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.
It is not true. Think for example at : $$x\mapsto\exp(-x)\sin(x)$$
It's still not true :-/
Consider the map :
$$f:\mathbb{R}\to\mathbb{R},x\mapsto\exp(-x)\left[\frac{1}{2}\left(\sin(x)-\cos(x)\right)-1\right]$$
Clearly, $f$ has zero limit at $+\infty$.
We have for all $x\in\mathbb{R}$ :
$$f'(x)=\exp(-x)\left(1+\cos(x)\right)\ge0$$
so that $f$ is increasing.
Now, for all $x\in\mathbb{R}$ :
$$f''(x)=-\exp(-x)\left(1+\sin(x)+\cos(x)\right)=-\exp(-x)\left(1+\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\right)$$
Hence, there is no $k\in\mathbb{R}$ such that $f$ would be concave on $(k,+\infty)$.
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 $n.$