Let $f:\mathbb{R} \to \mathbb{R}$ be a smooth function. I just answered a question of the form (I don't think the details are important):
Assuming that $\int_{1}^{\infty} f(x)/x\, dx$ exists, show [...]
I am able to prove:
Assuming $\lim_{x \to \infty} f(x) = 0$, [...]
Now, I suspect that $\int f(x)/x$ existing implies that $f(x) \to 0$. Clearly, if $f(x) \to L$, then $L=0$. But I can't see why the limit should exist. I expected some kind of integration by parts juggling to give something helpful, but I couldn't think of anything clever. I suspect the integrals $\int f(x)/x$ and $\int f'(x)$ are related in some relatively general way (by looking at polynomials as a case study), but I don't know exactly what it is.
I also suspect that $\int f(x)/x$ existing should mean $f(x) = O(x^{-k})$ with $k>0$, but I don't know how to make this precise either.
No, existence of $\int_1^\infty f(x)/x\,dx$ does not imply that $f(x)$ has a limit at $+\infty$.
If we're talking about improper Riemann integrals then $f(x)=\sin(x)$ is a counterexample. For the Lebesgue integral, let $$f=\sum_{n=1}^\infty\chi_{[n,n+1/n]}.$$