Proving $\lim\limits_{x \to \infty} xf(x)=0$ if $\int_{0}^{\infty}f(x) dx$ converges. Let $f(x)$ be a monotone non-increasing function such that $\int_{0}^{\infty}f(x) dx$ converges. Prove: $\lim\limits_{x \to \infty} xf(x)=0$. My question is, why can't I simply contradict any other possibility by using the Integral Limit Comparison Test with $1\over x$? I am, after all, to show that $f(x)=o(x)$ as $x\to \infty$. I really don't understand why monotony is crucial here. I could use some help.

Let
$F(x)
=\int_0^x f(t) dt
$.
We are given that
$\lim_{x \to \infty} F(x)$
exists.
Call this limit $L$.
Suppose it is not true that
$\lim_{x \to \infty} xf(x)
= 0
$.
Then there is a $c > 0$
such that
$x f(x) > c$
for arbitrarily large $x$.
Suppose
$x_0 f(x_0) > c$
for $x_0$ large enough that
$\int_{x_0}^{\infty} f(t) dt
< d
$.
Choose an $x_1 > x_0$ such that
$x_1 f(x_1) > c$.
Since $f$ is monotone decreasing,
$\int_{x_0}^{x_1} f(t) dt
> (x_1-x_0)f(x_1)
> (x_1-x_0)(c/x_1)
= c(1-x_0/x_1)
$.
If we choose
$x_1 > 2 x_0$,
then
$\int_{x_0}^{x_1} f(t) dt
> c/2
$.
So we have
$d
>\int_{x_0}^{\infty} f(t) dt
>\int_{x_0}^{x_1} f(t) dt
>c/2
$.
But since
$\lim_{x \to \infty} \int_{x}^{\infty} f(t) dt
=0
$,
we can choose $x_0$ large enough
so that $d$ is arbitrarily small.
In particular,
we can choose $d < c/2$,
which contradicts the
inequality above.
Therefore,
$\lim_{x \to \infty} xf(x)
= 0
$.
