TITLE: Prove, or disprove, $\int_{1}^{\infty} |f(x)|dx$ converge $\Longrightarrow \lim_{x\to\infty} f(x)/x=0$
QUESTION [0 upvotes]: $f(x)$ is a continuous function in $[1, \infty)$. I know that $\int_{1}^{\infty} |f(x)|dx 
$ converge $\Longrightarrow \int_{1}^{\infty} f(x)dx$ converge, so assuming $f(x)\ge0$ for $x\ge1$, $\lim_{x\to\infty} f(x)/x$ exists and because $\int_{1}^{\infty}xdx$ doesn't converge:

if $\lim_{x\to\infty} f(x)/x=L (L\in \mathbb{R})$ then $\int_{1}^{\infty} f(x)dx$ converge if and only if $\int_{1}^{\infty}xdx$ converge and that is a contradiction 
if $\lim_{x\to\infty} f(x)/x=\infty$ then if $\int_{1}^{\infty}xdx$ doesn't converge then $\int_{1}^{\infty} f(x)dx$ doesn't converge and that is a contradiction

so I'm left with $\lim_{x\to\infty} f(x)/x=0$. But what if $\exists x:x\ge1, f(t)<0$ for $t>x$ or $\lim_{x\to\infty} f(x)/x$ doesn't exists?

REPLY [0 votes]: This is not true. There is a continuous function $f$ such that $f(x)=n$ when $2n-\frac  1 {n^{3}} \leq x \leq 2n+\frac  1 {n^{3}}$ for each $n$ and $f$ is integrable on $[1,\infty)$. Note that $\frac {f(2n)} {2n} \to \frac  1  2$.
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