Let $\displaystyle f:[a,+\infty)\to\mathbb{R}$ be twice differentiable. If $\displaystyle \displaystyle\lim_{x\to\+\infty}{f(x)}=f(a)$ then there exists $\displaystyle x\in (a,+\infty)$ such that $\displaystyle f''(x)=0.$

This is what I have so far:

Suppose f is twice differentiable. Well, there are three cases:

(1) $\displaystyle \exists x_0, x_1$ such that $\displaystyle f''(x_0)< 0$ and $\displaystyle f''(x_1)> 0$; or

(2) $\displaystyle f''(x)\geq 0, \forall x$; or

(3) $\displaystyle f''(x)\leq 0, \forall x$.

In the first case, the conclusion follows immediately from Darboux's theorem. The second case implies that $\displaystyle f$ is convex. And I'm stuck here. One idea is to try to prove that $\displaystyle \exists x_0, x_1$ such that $\displaystyle f''(x_0)< 0$ and $\displaystyle f''(x_1)> 0$ and use Darboux's theorem again (assuming $\displaystyle f$ is not constant). But I don't know how to do it. I don't know if this (attempted) solution seems too complicated. Maybe there is an easier way. I'm open to suggestions.