Decreasing $f:\mathbb{R}\to\mathbb{R}$ tending to $0$ at $∞$ not convex beyond any point?

Question

Given a function $f:\mathbb{R}\to\mathbb{R}$ differentiable and strictly decreasing such that $\displaystyle \lim_{x\to\infty}f(x)=0$, I am looking to find out whether or not there exists an $x_0$ such that $f$ is convex on $(x_0,\infty)$. My guess is that the statement is not true but I can't find a counterexample.

Answer 1

Edit: added additional term to make the function actually decreasing. The idea remains the same.

Consider something like $$\frac1x + \frac{\sin(x)}{x^2}+ \frac{4 \sin^2(x/2)}{x^3}.$$ Notice how it keeps wiggling all the way down. Here is its graph from $100$ to $120$ to give an impression.