As a physicist, I tend to think about $\sim 1/x$ as the "slowest" fall-off of a "reasonable" function. Let us state this formally: $${\rm lim}_{x \to \infty} f(x) = 0, f(x) \in Reas \implies \exists A \in \mathbb{R}: {\rm lim}_{x \to \infty} x f(x) = A \,. $$ Where $Reas$ can be for example the set of real analytic functions $C^\omega (\mathbb{R})$.
For the case of strictly real analytic functions I have a hunch that the $0$ infinity limit could even imply quicker fall-offs, so let us think about $C^\infty(\mathbb{R})$ where things such as $x/(x^2+ \epsilon)$ definitely have the $1/x$ asymptotics.
So, is there a class of functions $SloFall\subset C^\infty(\mathbb{R})$ for which
$$\forall f(x) \in SloFall: {\rm lim}_{x \to \infty} f(x) = 0 \land {\rm lim}_{x \to \infty} |x f(x)| = \infty \,? $$
Could you give some examples of such functions? If such functions do not exist in $C^\infty(\mathbb{R})$, would there be an example of such a function say in at most $C^1(\mathbb{R})$?
There are innumerable examples, and you needn't use transcendentals like logarithms. On the interval $[1,\infty]$ let $f(x)$ be the reciprocal of the square root or the cube root or ... etc. These fall off more slowly than $1/x$ but $xf(x)$ becomes infinite. Shift the domain left if you want to incorporate values less than 1. In a more theoretical mode, you can smoothly splice a function onto one of these at the left endpoint to demonstrate the existence of functions in $C^\infty(\mathbb{R})$ that have the same asymptotic behavior.