Zero and Infinite Limits in Comparison Tests

Why a Ratio Limit Gives a One-Way Comparison

When two nonnegative integrands have a ratio tending to zero, the numerator is eventually smaller than the denominator. When the ratio tends to infinity, the numerator is eventually larger. These eventual inequalities give useful conclusions about improper integrals, but they do not make the integrals equivalent: in each case, some convergence questions remain unanswered.

The direction of the inequality is decisive. A smaller nonnegative integrand is easier to integrate, so convergence of a larger integrand can force convergence of the smaller one. But divergence of a larger integrand does not, by itself, force divergence of a smaller one. The reverse situation applies when the numerator eventually dominates.

Setting and Definitions

Consider nonnegative functions \(f\) and \(g\) on \((a,b]\), where \(a\) is an endpoint at which the integrals may be improper. Assume that both functions are integrable on every closed interval \([c,b]\) with \(a<c<b\). Their improper integrals at \(a\) are defined by

\int_a^b f(x)\,dx = \lim_{\varepsilon\to 0^+} \int_{a+\varepsilon}^b f(x)\,dx, \qquad \int_a^b g(x)\,dx = \lim_{\varepsilon\to 0^+} \int_{a+\varepsilon}^b g(x)\,dx.

Because the integrands are nonnegative, each limit exists either as a finite number or as \(+\infty\). The improper integral converges if its limit is finite and diverges if the limit is \(+\infty\).

Definition: The ratio comparison examines \(f(x)/g(x)\) near the improper endpoint. Assume \(g(x)>0\) sufficiently close to \(a\), so this ratio is defined there. A limit of zero means that \(f\) is eventually smaller than \(g\) by an arbitrarily strong factor; a limit of \(+\infty\) means that \(f\) is eventually larger than \(g\) by an arbitrarily strong factor.

The functions need only satisfy the ratio condition near \(a\). Any portion of the integrals away from \(a\) is finite under the stated local integrability assumption and cannot change whether the improper integral converges or diverges.

Ratio Tending to Zero

Suppose \(f(x)/g(x)\to 0\) as \(x\to a^+\). In particular, the ratio is eventually less than \(1\). Therefore, for some \(c>a\),

0\le f(x)\le g(x) \qquad\text{for }a<x<c.

This is the key inequality. It gives the following two implications:

Theorem: If \(f,g\ge 0\) and \(f(x)/g(x)\to 0\) at the improper endpoint, then:

if \(\int_a^b g(x)\,dx\) converges, then \(\int_a^b f(x)\,dx\) converges;
if \(\int_a^b f(x)\,dx\) diverges, then \(\int_a^b g(x)\,dx\) diverges.

The second statement is the contrapositive of the first. Both express the same ordering: the smaller integrand cannot have a larger integral than the larger one on the region where \(f\le g\). In particular, convergence of the larger integral guarantees convergence of the smaller, while divergence of the smaller forces divergence of the larger.

Why the Implication Holds

Choose \(c\) close enough to \(a\) that \(f(x)\le g(x)\) on \((a,c)\). For each \(\varepsilon>0\) small enough, integrate this inequality from \(a+\varepsilon\) to \(c\):

0\le \int_{a+\varepsilon}^{c} f(x)\,dx \le \int_{a+\varepsilon}^{c} g(x)\,dx.

If the improper integral of \(g\) converges, the right-hand side stays bounded as \(\varepsilon\to0^+\). The nonnegative integrals of \(f\) on these truncated intervals are therefore bounded as well, so their increasing limit is finite. Thus the integral of \(f\) converges. Conversely, if the integral of \(f\) grows without bound, the inequality means the integral of \(g\) must also grow without bound. Adding the finite portions from \(c\) to \(b\) does not affect either conclusion.

Worked Example: A Convergent Numerator Compared with a Divergent Denominator

On \((0,1]\), let \(f(x)=1/\sqrt{x}\) and \(g(x)=1/x\). Both are nonnegative, and

\frac{f(x)}{g(x)} = \frac{1/\sqrt{x}}{1/x} = \sqrt{x} \longrightarrow 0 \qquad (x\to0^+).

The improper integral of \(f\) is

\int_0^1 \frac{1}{\sqrt{x}}\,dx = \lim_{\varepsilon\to0^+} \int_\varepsilon^1 x^{-1/2}\,dx = \lim_{\varepsilon\to0^+} \left[2\sqrt{x}\right]_\varepsilon^1 = \lim_{\varepsilon\to0^+}(2-2\sqrt{\varepsilon}) = 2.

The antiderivative checks because \(\frac{d}{dx}(2\sqrt{x})=1/\sqrt{x}\) for \(x>0\). The integral of \(g\), in contrast, is

\int_0^1 \frac{1}{x}\,dx = \lim_{\varepsilon\to0^+} \int_\varepsilon^1 \frac{1}{x}\,dx = \lim_{\varepsilon\to0^+} \left[\ln x\right]_\varepsilon^1 = \lim_{\varepsilon\to0^+}(-\ln\varepsilon) = +\infty.

Here \(\frac{d}{dx}\ln x=1/x\) for \(x>0\). The result is consistent with the theorem: the larger integral diverges, while the smaller one converges. It also shows that divergence of \(g\) does not decide whether \(f\) converges.

Ratio Tending to Infinity

Now suppose \(f(x)/g(x)\to+\infty\) as \(x\to a^+\). The ratio is eventually greater than \(1\), so for some \(c>a\),

0\le g(x)\le f(x) \qquad\text{for }a<x<c.

The roles of the two functions are reversed. Integrating the inequality gives the corresponding implications:

Theorem: If \(f,g\ge 0\) and \(f(x)/g(x)\to+\infty\) at the improper endpoint, then:

if \(\int_a^b f(x)\,dx\) converges, then \(\int_a^b g(x)\,dx\) converges;
if \(\int_a^b g(x)\,dx\) diverges, then \(\int_a^b f(x)\,dx\) diverges.

In this case \(f\) is eventually the larger function. Convergence of its integral forces convergence of the smaller integral, and divergence of the smaller integral forces divergence of the larger one.

Worked Example: A Divergent Numerator Compared with a Convergent Denominator

On \((0,1]\), let \(f(x)=1/x\) and \(g(x)=1\). Their ratio satisfies

\frac{f(x)}{g(x)} = \frac{1/x}{1} = \frac{1}{x} \longrightarrow+\infty \qquad (x\to0^+).

The integral of \(f\) is

\int_0^1 \frac{1}{x}\,dx = \lim_{\varepsilon\to0^+} \left[\ln x\right]_\varepsilon^1 = \lim_{\varepsilon\to0^+}(-\ln\varepsilon) = +\infty.

Again, \(\ln x\) is an antiderivative because its derivative is \(1/x\). The integral of \(g\) is

\int_0^1 1\,dx = \lim_{\varepsilon\to0^+} \left[x\right]_\varepsilon^1 = \lim_{\varepsilon\to0^+}(1-\varepsilon) = 1.

The antiderivative \(x\) checks since its derivative is \(1\). Thus the larger integral diverges and the smaller one converges. The example confirms that divergence of the larger integral does not force divergence of the smaller one.

What Each Ratio Limit Leaves Undecided

A zero or infinite ratio gives one-sided implications, not a complete decision in every situation. The unaddressed cases are not technical gaps: examples show that either outcome can occur.

Ratio limit	Conclusion that is guaranteed	Question left undecided
\(f/g\to0\)	\(\int g\) converges \(\Rightarrow\) \(\int f\) converges; \(\int f\) diverges \(\Rightarrow\) \(\int g\) diverges	If \(\int g\) diverges, whether \(\int f\) converges or diverges
\(f/g\to+\infty\)	\(\int f\) converges \(\Rightarrow\) \(\int g\) converges; \(\int g\) diverges \(\Rightarrow\) \(\int f\) diverges	If \(\int f\) diverges, whether \(\int g\) converges or diverges

For a ratio tending to zero, the first worked example already shows that the larger integral can diverge while the smaller one converges. The next example shows that the smaller integral can instead diverge too. For a ratio tending to infinity, the second worked example shows that the larger integral can diverge while the smaller one converges. The final example shows that the smaller integral can converge while the larger one also converges.

Worked Example: A Zero Ratio with Both Integrals Divergent

Let \(f(x)=1/x\) and \(g(x)=1/x^2\) on \((0,1]\). Then

\frac{f(x)}{g(x)} = \frac{1/x}{1/x^2} = x \longrightarrow 0 \qquad (x\to0^+).

For \(f\),

\int_0^1 \frac{1}{x}\,dx = \lim_{\varepsilon\to0^+} \left[\ln x\right]_\varepsilon^1 = \lim_{\varepsilon\to0^+}(-\ln\varepsilon) = +\infty.

For \(g\),

\int_0^1 \frac{1}{x^2}\,dx = \lim_{\varepsilon\to0^+} \left[-\frac{1}{x}\right]_\varepsilon^1 = \lim_{\varepsilon\to0^+} \left(-1+\frac{1}{\varepsilon}\right) = +\infty.

The antiderivatives check: \(\frac{d}{dx}\ln x=1/x\) and \(\frac{d}{dx}(-1/x)=1/x^2\) for \(x>0\). Thus a zero ratio does not imply that the numerator integral converges. It permits both convergence and divergence of that integral when the denominator integral diverges.

Worked Example: An Infinite Ratio with Both Integrals Convergent

Let \(f(x)=x^{-1/2}\) and \(g(x)=x^{-1/4}\) on \((0,1]\). The ratio is

\frac{f(x)}{g(x)} = x^{-1/2-(-1/4)} = x^{-1/4} \longrightarrow+\infty \qquad (x\to0^+).

For \(f\),

\int_0^1 x^{-1/2}\,dx = \lim_{\varepsilon\to0^+} \left[2x^{1/2}\right]_\varepsilon^1 = \lim_{\varepsilon\to0^+}(2-2\sqrt{\varepsilon}) = 2.

For \(g\),

\int_0^1 x^{-1/4}\,dx = \lim_{\varepsilon\to0^+} \left[\frac{4}{3}x^{3/4}\right]_\varepsilon^1 = \lim_{\varepsilon\to0^+} \left(\frac{4}{3}-\frac{4}{3}\varepsilon^{3/4}\right) = \frac{4}{3}.

The derivatives of the displayed antiderivatives are \(x^{-1/2}\) and \(x^{-1/4}\), respectively. Both integrals converge even though \(f/g\to+\infty\). Therefore, convergence of the smaller integral \(g\) does not determine whether the larger integral \(f\) converges.

A Common Direction Error

A frequent mistake is to infer that a divergent larger integral forces a smaller integral to diverge. That inference reverses the valid comparison. If \(0\le f\le g\), then convergence of \(g\) controls \(f\), but divergence of \(g\) does not control \(f\).

Incorrect reasoning: Since \(f/g\to0\), we have \(f\le g\) near the endpoint. If \(\int g\) diverges, then \(\int f\) must diverge too. Correction: The inequality \(f\le g\) gives the implication from convergence of \(g\) to convergence of \(f\), not from divergence of \(g\) to divergence of \(f\). For example, \(f(x)=1/\sqrt{x}\) and \(g(x)=1/x\) have \(f/g\to0\), but \(\int_0^1 f(x)\,dx=2\) while \(\int_0^1 g(x)\,dx=+\infty\).

A reliable check is to write the eventual inequality before stating any conclusion. Then apply the direct comparison rule in the direction that inequality permits. For nonnegative integrands, an upper bound can establish convergence; a lower bound can establish divergence.

Using the Test Carefully

Before applying either ratio conclusion, verify three points:

The integrands are nonnegative near the improper endpoint.
The denominator \(g(x)\) is positive sufficiently close to the endpoint, so the ratio is defined.
The ratio tends to zero or \(+\infty\), and the resulting eventual inequality is in the direction needed for the desired conclusion.

If the ratio tends to zero, use the test to transfer convergence from \(g\) to \(f\), or divergence from \(f\) to \(g\). If it tends to \(+\infty\), transfer convergence from \(f\) to \(g\), or divergence from \(g\) to \(f\). In the other cases, the ratio alone does not settle the question; another comparison or a direct evaluation may be needed.

The essential lesson is not just to compute a limit, but to keep track of which function is eventually larger. That ordering determines exactly what the comparison can prove—and where its information ends.

Check Your Understanding

For each question, identify the eventual inequality before deciding what the ratio test guarantees.

If \(f/g\to0\) and \(\int_a^b g(x)\,dx\) converges, what can be concluded about \(\int_a^b f(x)\,dx\)?
If \(f/g\to0\) and \(\int_a^b g(x)\,dx\) diverges, must \(\int_a^b f(x)\,dx\) diverge? Explain.
If \(f/g\to+\infty\) and \(\int_a^b g(x)\,dx\) diverges, what can be concluded about \(\int_a^b f(x)\,dx\)?
If \(f/g\to+\infty\) and \(\int_a^b f(x)\,dx\) diverges, does the ratio determine whether \(\int_a^b g(x)\,dx\) converges?
What eventual inequality follows from \(f/g\to0\), and which direction of comparison conclusion does it support?