Finite limit of $f/g$ at infinity implies integrable Let $f,g : [a,\infty)\to \mathbb{R}$ be such that $f/g$ has finite limit at $\infty$.
If either $f$ or $g$ is Lebesgue-integrable then why must the other be also?
I know I have to use the comparison test but I’m not sure how.
$a=1, f(x)=\frac 1 {x^{2}}, g(x) =\frac 1 x$ is a counter-example. However, if the limit of $\frac f g$ is finite and nonzero then the conclusion holds.
Suppose $\displaystyle \lim_{x\,\to\,+\infty} \frac {f(x)} {g(x)} = L \in (0,+\infty)$ and $\displaystyle\int\limits_{[a,+\infty)} |g(x)|\,dx &lt;+\infty.$ Then for some positive number $b$ you have $0 &lt; f(x) &lt; (L+1)g(x)$ whenever $x&gt;b.$ Therefore
$$
\int\limits_{[a,+\infty)} |f(x)|\,dx \le \int\limits_{[a,+\infty)} (L+1)|g(x)|\,dx &lt; +\infty.
$$
Since we assumed $L\ne0$ above, you can apply the same argument with $g/f$ in place of $f/g$ and $1/L$ in place of $L.$ But that doesn't work if $L=0.$
But what if $x\not&gt;b\text{ ?}$ We can't say what happens if $a&lt;x\le b.$