Let $f:[0,\infty)\to\mathbb{R}$ be integrable in everywhere.
Suppose $\int\limits_0^{\infty}|f(t)|dt$ converges. Show that there exists a sequence $x_n$ such that $x_n\to\infty$ while $f(x_n)\to 0$.
Show that if we require $f$ to be continuous, then the above holds even if $\int\limits_0^{\infty}f(t)dt$ converges (not necessarily absolutely).
Is the statement in (1) true if we drop the requirement for continuity?
For (1), suppose every sequence such that $x_n\to\infty$, $f(x_n)$ does not tend to $0$. Then necessarily there exists $L>0$ such that from some point $x_0$, for all $x>x_0$, $|f(x)| \geq L$. But then
$$
|f(x)| > L \Rightarrow \int\limits_{x_0}^\infty |f(t)|dt \geq \int\limits_{x_0}^{\infty}Ldx = \infty
$$
a contradiction for the convergence of the improper integral.
For (2), I am not really sure. The only conclusion I got is that if $f$ is continuous and there is no such sequence $x_n$, then necessarily $f$ intersects the $x$ axis only finitely many times.
Edit:
For (3) it is proved here.
Best Answer
For (2) by the same reasoning by contradiction, there exist $L>0$ and $x_0$ such that for all $x>x_0$, $|f(x)| \geq L$. Because $f$ is continuous, either $f(x) \geq L$ for all $x>x_0$ or $f(x) \leq -L$ for all $x>x_0$.
Then the conclusion is similar than (1) (distinguish the two cases).
For (3) I think the counter-example you provide actually works.