What an Unbounded Integrand Can Do Near a Singularity
An integrand may grow without bound as its input approaches a singular point. But “unbounded” does not specify how its values behave. They may eventually become positive and exceed every fixed bound, eventually become negative and fall below every fixed bound, or keep changing in a way that prevents any one-sided limit.
The distinction matters when describing a singularity and when deciding what a limit calculation actually establishes. In particular, seeing large values is not enough to conclude that a function tends to positive or negative infinity. The sign and the behavior throughout a whole one-sided neighborhood must be considered.
One-Sided Limits to Signed Infinity
Let \(a\) be a finite point, and suppose \(f\) is defined for \(x\) in some interval immediately to the right of \(a\). We say that \(f(x)\) tends to positive infinity as \(x\) approaches \(a\) from the right if, for every real number \(M\), there is a \(\delta>0\) such that
We write \(\lim_{x\to a^+}f(x)=+\infty\). The definition for a left-hand limit uses \(0<a-x<\delta\) instead. For a limit of negative infinity, the corresponding condition is \(f(x)<M\) for every chosen real bound \(M\), sufficiently close to \(a\) on the specified side.
These are statements about eventual behavior, not about a few selected values. For example, a function that takes arbitrarily large positive values near \(a\) is unbounded above there, but it need not tend to positive infinity: it might also take arbitrarily large negative values arbitrarily close to \(a\).
Reading the Sign Near a Pole
For many rational expressions, the sign near a singularity can be found by separating the numerator from the power in the denominator. Suppose \(g\) is continuous near \(a\) and \(g(a)\ne 0\), and consider
where \(m\) is a positive integer. Continuity ensures that sufficiently close to \(a\), \(g(x)\) has the same sign as \(g(a)\) and stays bounded away from zero. The denominator’s sign depends on the parity of \(m\): an even power is positive on both sides, while an odd power has the sign of \(x-a\). Meanwhile, \(\lvert x-a\rvert^m\) tends to zero. These facts together determine the signed infinite limit on each side.
If it is continuous and nonzero at the singular point, record its sign there. That sign persists sufficiently close to the point.
For an even power it is positive on both sides; for an odd power it changes sign across the point.
The quotient’s magnitude grows without bound, and its eventual sign identifies whether the limit is positive or negative infinity.
The sign analysis must be done separately on each side. A two-sided limit to a single signed infinity exists only when both one-sided limits have that same behavior. If the left and right limits have opposite signs, there is no two-sided limit, even though each one-sided limit exists in the extended sense.
Worked Example: An Even-Order Pole
Find the one-sided limits of \(f(x)=\frac{2x+1}{(x-3)^2}\) as \(x\) approaches \(3\).
The numerator tends to \(2(3)+1=7\), which is positive. More explicitly, if \(x\) is close enough to \(3\), then \(2x+1>0\). The denominator is a square, so \((x-3)^2>0\) for every \(x\ne 3\). Thus \(f(x)>0\) on both sides near \(3\).
For any \(M>0\), choose \(\delta>0\) small enough that \(2x+1>6\) whenever \(0<|x-3|<\delta\), and also choose \(\delta<\sqrt{6/M}\). Then
Therefore the function tends to positive infinity from either side:
The sign and magnitude agree with the conclusion: the numerator stays positive, and the positive denominator approaches zero. This also verifies why the even power produces the same one-sided behavior.
Worked Example: An Odd-Order Pole
Find the one-sided limits of \(f(x)=\frac{x+2}{(x-1)^3}\) as \(x\) approaches \(1\).
Near \(1\), the numerator is positive: \(x+2\) tends to \(3\). The denominator has an odd power. If \(x<1\), then \((x-1)^3<0\), so the quotient is negative. If \(x>1\), then \((x-1)^3>0\), so the quotient is positive.
For the left-hand limit, the positive numerator stays bounded away from zero while the negative denominator approaches zero. Hence the quotient falls below every real bound. For the right-hand limit, it exceeds every real bound. Thus
The two one-sided limits are different, so there is no two-sided limit at \(1\). The calculation is checked by the signs on each side: negative on the left, positive on the right, with magnitude growing without bound on both.
Unboundedness Does Not Guarantee a Signed Limit
The rational examples have a stable sign on each side of the singularity. Other functions can change sign repeatedly as they approach the singular point. To rule out a proposed limit, it is often useful to choose two sequences approaching the point along which the function has incompatible behavior. If a one-sided limit exists, the function values along every sequence approaching from that side must approach the same limit, including the same signed infinity when that is the claimed limit.
Worked Example: An Unbounded Oscillating Integrand
Consider \(f(x)=\frac{\sin(1/x)}{x^2}\) as \(x\to0^+\). We will show that the function is unbounded near zero but has no one-sided limit, even as a signed infinity.
For positive integers \(n\), set
Both sequences are positive and tend to zero. Their sine values are
Consequently,
whereas
Thus the function takes arbitrarily large positive and negative values arbitrarily close to zero. It cannot tend to positive infinity, because the values along \(y_n\) do not eventually exceed every bound. It cannot tend to negative infinity, because the values along \(x_n\) do not eventually fall below every bound. It has no finite limit either, since these sequences do not approach the same finite value. Therefore \(\lim_{x\to0^+}f(x)\) does not exist.
This confirms both parts of the conclusion: the function is unbounded near zero, but its sign oscillation prevents a one-sided limit.
A Limit of the Integrand Is Not a Limit of Its Integral
In an improper integral, it is not the integrand’s one-sided limit that defines convergence. The relevant limit is the limit of the accumulated integral as the endpoint approaches the singularity. The two limits answer different questions. An integrand can tend to positive infinity while its improper integral converges, or an integrand can fail to have a limit while the accumulated integral has its own behavior.
Worked Example: An Infinite Integrand With a Finite Improper Integral
Consider \(\int_0^1 \frac{1}{\sqrt{x}}\,dx\). The integrand tends to positive infinity as \(x\to0^+\), since \(1/\sqrt{x}\) grows without bound. To evaluate the improper integral, first keep the lower endpoint positive:
Now take the endpoint limit:
Therefore the improper integral converges to \(2\), even though its integrand tends to positive infinity. Differentiating the antiderivative checks the calculation: \(\frac{d}{dx}(2\sqrt{x})=1/\sqrt{x}\) for \(x>0\). The example shows why a signed infinite limit of the integrand should not be confused with divergence of the integral.
Common Mistake: Treating “Unbounded” as “Tends to Infinity”
A tempting but invalid argument is: “The denominator tends to zero, so the quotient tends to positive infinity.” This omits the numerator’s sign and whether the denominator changes sign. It also fails for oscillating numerators. For the function \(\sin(1/x)/x^2\), the denominator is positive and tends to zero, but the numerator alternates between \(1\) and \(-1\) along sequences approaching zero. The conclusion \(+\infty\) is therefore false.
The correction is to establish eventual behavior, not just large magnitude. For a signed infinite limit, the function must eventually have the appropriate sign and exceed every bound in magnitude on the entire sufficiently small one-sided interval. If positive and negative values persist arbitrarily close to the point, neither signed infinite limit is possible.
Why This Distinction Matters
When studying a singular integrand, identify the side of approach before deciding what a limit says. An even power in a pole may give the same sign on both sides; an odd power may give opposite signs. Oscillation can prevent a signed limit even when values grow arbitrarily large in magnitude. These are distinct behaviors and should not all be described simply as “blowing up.”
Keep the object being limited explicit. The statement \(\lim_{x\to a^+}f(x)=+\infty\) describes the values of the integrand. Convergence of an improper integral is instead determined by the limit of an integral with its endpoint held away from the singularity. A correct analysis identifies which limit is being evaluated and checks the necessary behavior on the relevant side.
Check Your Understanding
For each question, distinguish carefully between unboundedness, a signed infinite limit, and failure of a limit.
- For \(f(x)=\frac{1}{(x-2)^4}\), what are the one-sided limits as \(x\to2\)? Explain the role of the even power.
- For \(g(x)=\frac{5}{(x+1)^3}\), determine the left- and right-hand limits as \(x\to-1\).
- What must be true of \(f(x)\) sufficiently close to \(a\) for \(\lim_{x\to a^+}f(x)=+\infty\)?
- Why does showing that a function takes arbitrarily large positive values near a point not prove that it tends to positive infinity?
- What two sequences could be used to test whether \(\frac{\sin(1/x)}{x^2}\) has a one-sided limit as \(x\to0^+\)?
- Why does an integrand tending to positive infinity not, by itself, establish that its improper integral diverges?
