# Limit comparison test for Improper Integrals

The test says:

Given two functions $f$ and $g$ defined positive and integrable on $\left [ a,+\infty \right [$. Suppose that $\lim_{x\to \infty}\frac{f(x)}{g(x)}=L$ exists.

If $0< L < \infty$: $\int_{a}^{\infty }g(x)dx$ converges $\Leftrightarrow \int_{a}^{\infty }f(x)dx$ converges

If $L=0$: $\int_{a}^{\infty }g(x)dx$ converges $\Rightarrow \int_{a}^{\infty }f(x)dx$ converges

The proof goes by using the definition of the limit at infinity by setting an $\epsilon$ and stating the existence of a certain $M$ above which $\frac{f(x)}{g(x)}$ is within $\epsilon$ of $L$ and then using the comparison theorem to deduce the convergence of the integral of a function by its upper boundedness by a function whose integral converges. Here is a link if I explained badly (from the end of page 1 to the top of page 2): http://www.math.toronto.edu/~alfonso/137/137_1516_LCT.pdf

MY QUESTION:

I fail to see why this proof fails in the left implication direction when $L=0$? We would have $-\epsilon g(x)\leq f(x)\leq \epsilon g(x)$ so why would convergence of $\int_{a}^{\infty }f(x)dx$ not imply convergence of $\int_{a}^{\infty }g(x)dx$?

• Notice that the Limit comparison test does not assume that $0\le f(x)\le g(x)$, and that the first statement should say if $0<L<\infty$, then $\int_a^{\infty}f(x)dx$ converges $\iff$ $\int_a^{\infty}g(x)dx$ converges. – user84413 Aug 28 '16 at 22:14
A basic counterexample would be $f(x)=\frac 1{x^2}$ and $g(x)=\frac 1x$. $\lim_{x\to\inf}\frac{f(x)}{g(x)}=0$, and $\int_{a>0}^\infty f(x)dx$ converges, but $\int_{a>0}^\infty g(x)dx$ does not. The reason the proof breaks down can be understood by looking at the inequality you gave.
$$x>M\implies-\epsilon g(x)\le f(x)\le\epsilon g(x)$$
The right inequality is a perfect fine for comparison test, since $0\le f(x)\le \epsilon g(x)$, but since the $f$, $g$, and $\epsilon$ are nonegative, the left side is trivially true, and is useless for a comparison. More specifically, the inequality $-\epsilon g(x)\le 0\le f(x)$ does not squeeze $g(x)$ between two functions with the same limit. That is precisely why the proof uses $\epsilon=\frac L2$, assuring that the left inequality gives the useful comparison $\frac L2g(x)\le f(x)\le 0$. When $L=0$, the argument no longer works. In fact, no argument would work, as this would contradict the counterexamples.
• +1 Perfect explanation! Minor note, I think you may have swapped $f(x)$ and g(x)$while writing the explanation so I corrected it. – John11 Aug 28 '16 at 22:47 • Thanks for pointing that out. was thinking$\ge$but writing$\le$. – Kajelad Aug 28 '16 at 23:02 Let$f(x)=\frac{1}{x^2}$and$g(x)=\frac{1}{x}$. Then$\displaystyle\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty}\frac{1/x^2}{1/x}=\lim_{x\to\infty}\frac{1}{x}=0$and$\int_1^{\infty}f(x)dx$converges, but$\int_1^{\infty}g(x)dx\$ diverges.