Introduction
A rational function can become unbounded near an input where its denominator is zero. The direction of that divergence depends on the denominator’s sign on each side of the input. A denominator factor with an even power keeps the same sign on both sides; a factor with an odd power changes sign. This parity difference determines whether the two one-sided limits agree or have opposite signs.
The numerator also matters: its sign near the input can make both limits positive or both negative, or determine which side tends to positive infinity for an odd-order pole. We will make these statements precise and prove the sign rule.
Denominator Zero Order and the Local Form
Let \(f(x)=P(x)/Q(x)\), where \(P\) and \(Q\) are polynomials and \(Q\) is not the zero polynomial. Suppose \(Q(a)=0\). The multiplicity, or order, of this zero is the positive integer \(m\) for which we can write
Thus the factor \(x-a\) occurs exactly \(m\) times in the factorization of \(Q\). In this tutorial we focus on the case \(P(a)\ne 0\). Then no factor of the numerator vanishes at \(a\), and the rational function has a genuine pole there of order \(m\). The function is undefined at \(a\), but it is defined at sufficiently nearby inputs other than \(a\).
For \(x\ne a\), write the function in the form
Because \(G(a)\ne 0\), the quotient \(H=P/G\) is defined and continuous near \(a\). Its value at \(a\) is the nonzero number
Continuity means that sufficiently close to \(a\), \(H(x)\) has the same sign as \(C\). The factor \((x-a)^m\) supplies the remaining sign change, if there is one.
The Parity Sign Rule
On the right of \(a\), \(x-a>0\), so \((x-a)^m>0\) for every positive integer \(m\). On the left, \(x-a<0\). Its \(m\)-th power is negative if \(m\) is odd and positive if \(m\) is even. Consequently, the sign of \(f(x)=H(x)/(x-a)^m\) near \(a\) follows this rule:
| Order \(m\) | Sign on the left of \(a\) | Sign on the right of \(a\) |
|---|---|---|
| Even | Same as \(C\) | Same as \(C\) |
| Odd | Opposite to \(C\) | Same as \(C\) |
Here, “has the sign of \(C\)” means the corresponding limit is \(+\infty\) when \(C>0\), and \(-\infty\) when \(C<0\). A sign opposite to \(C\) reverses those outcomes.
Why the Function’s Magnitude Grows Without Bound
The sign chart alone does not establish that the function’s magnitude actually becomes arbitrarily large. To show this, first use continuity of \(H\) at \(a\). Since \(H(a)=C\ne0\), there is a number \(\delta_0>0\) such that
The triangle inequality then gives \(|H(x)|>|C|/2\), and \(H(x)\) has the same sign as \(C\), whenever \(|x-a|<\delta_0\). Now fix any \(R>0\), the desired lower bound for the magnitude. Choose
If \(0<|x-a|<\delta\), then \(x\) lies in the neighborhood where \(|H(x)|>|C|/2\), and also \(|x-a|^m<|C|/(2R)\). Therefore
Because this holds for every \(R>0\), the magnitude of \(f(x)\) grows without bound as \(x\) approaches \(a\) from either side. Combining this magnitude result with the sign rule proves the stated one-sided limits. Notice that \(\delta\) must also keep \(x\) inside the neighborhood controlled by continuity; the minimum in its definition ensures that condition.
Worked Examples
Worked Example: An Even-Order Pole with Positive Limits
Find the one-sided limits of
at \(a=1\). The denominator has the factor \((x-1)^2\), so the order is \(m=2\), which is even. The remaining factor is \(G(x)=x+2\), and the numerator is \(P(x)=x^2+1\). Evaluate the local coefficient:
Thus \(H(x)=(x^2+1)/(x+2)\) is positive near \(1\). The denominator factor \((x-1)^2\) is positive on both sides, so \(f(x)\) is positive on both sides and its magnitude grows without bound. Hence
As a sign check, at \(x=0.9\) the numerator is \(1.81\) and the denominator is \(0.01(2.9)=0.029\), giving \(f(0.9)=1810/29>0\). At \(x=1.1\), the numerator is \(2.21\) and the denominator is \(0.01(3.1)=0.031\), giving \(f(1.1)=2210/31>0\). These checks agree with the one-sided signs; the unboundedness follows from the argument above, not from these sample values.
Worked Example: An Even-Order Pole with Negative Limits
Determine the one-sided limits at \(a=3\) for
The order is \(m=4\), which is even. Here \(P(x)=-(x+2)\), \(G(x)=x+1\), and
The even power \((x-3)^4\) is positive on both sides of \(3\), while \(H(x)=-(x+2)/(x+1)\) is negative near \(3\). Therefore \(f(x)\) is negative on both sides, and
For a numerical sign check, at \(x=2.9\) the numerator is \(-4.9\), and the denominator is \((-0.1)^4(3.9)=0.00039\), so \(f(2.9)=-490000/39<0\). At \(x=3.1\), the numerator is \(-5.1\), and the denominator is \((0.1)^4(4.1)=0.00041\), so \(f(3.1)=-510000/41<0\). Both values have the predicted negative sign.
Worked Example: An Odd-Order Pole with Opposite Limits
Find the one-sided limits at \(a=2\) for
The order is \(m=3\), which is odd. The remaining factor is \(G(x)=x+4\), so
On the right of \(2\), \((x-2)^3>0\), and \(H(x)\) is positive near \(2\). Thus \(f(x)\) is positive and grows without bound there. On the left, \((x-2)^3<0\), while \(H(x)\) remains positive, so \(f(x)\) is negative and its magnitude grows without bound. Therefore
For an exact sign check, at \(x=1.9\) the numerator is \(2.9\), and the denominator is \((-0.1)^3(5.9)=-0.0059\), so \(f(1.9)=-29000/59<0\). At \(x=2.1\) the numerator is \(3.1\), and the denominator is \((0.1)^3(6.1)=0.0061\), so \(f(2.1)=31000/61>0\). The signs match the predicted left-negative, right-positive pattern.
A Common Mistake
A frequent error is to assume that every pole has opposite signs on its two sides. That is true for an odd-order pole, but not for an even-order pole. For instance, consider \(g(x)=1/(x-2)^2\). An incorrect sign argument might treat the left side as though a first power of \(x-2\) were present and conclude
But when \(x<2\), \(x-2<0\) and \((x-2)^2>0\); when \(x>2\), \(x-2>0\) and again \((x-2)^2>0\). The numerator is positive on both sides. Thus the correct limits are
Always determine the sign of the full power \((x-a)^m\), not just the sign of \(x-a\).
Using the Rule Efficiently
For a rational function with a genuine pole at \(a\), the following sequence organizes the calculation:
Write the denominator as \(Q(x)=(x-a)^mG(x)\), where \(G(a)\ne0\).
Confirm that \(P(a)\ne0\), so the zero is not shared by the numerator in this setup.
Evaluate \(C=P(a)/G(a)\) to determine the sign of the remaining factor near \(a\).
Even \(m\) gives matching signs; odd \(m\) gives opposite signs. The right-hand sign is the sign of \(C\).
The condition \(P(a)\ne0\) matters. If the numerator also vanishes at \(a\), this pole-order rule cannot be applied directly in the form given here; the numerator’s vanishing must also be accounted for.
Check Your Understanding
For each question, identify the order and the sign of the local coefficient before deciding on the one-sided limits.
- For \(f(x)=2/(x+1)^4\), what are the limits as \(x\to-1^-\) and \(x\to-1^+\)?
- For \(g(x)=-3/(x-4)^3\), which side tends to positive infinity?
- If \(Q(x)=(x-a)^5G(x)\), \(G(a)<0\), and \(P(a)>0\), what are the signs of the left- and right-hand limits of \(P(x)/Q(x)\)?
- Why is it necessary to require \(P(a)\ne0\) for the pole-order rule stated here?
- What additional choice must be included in \(\delta\) when proving that the function’s magnitude exceeds any prescribed \(R>0\)?
