Why Multiplicity Determines the Sign Pattern
A rational function can become unbounded near a zero of its denominator. But knowing that it grows without bound is not enough: we also need to determine whether its values grow toward positive infinity or negative infinity on each side. The key is to identify the denominator’s effective multiplicity after cancellation and then track the sign of the remaining factors.
Multiplicity controls whether the sign reverses when we cross the point. An odd power of \(x-a\) changes sign from one side of \(a\) to the other; an even power does not. The other factors determine which infinity occurs on either side.
Multiplicity and Cancellation
A polynomial \(P\) has a zero of multiplicity \(r\) at \(a\) if it can be factored as \(P(x)=(x-a)^r p(x)\), where \(p(a)\ne 0\). The exponent \(r\) counts how many copies of the factor \(x-a\) occur. For example, \((x-2)^3(x+1)\) has a zero of multiplicity \(3\) at \(x=2\), since the remaining factor \(x+1\) is nonzero there.
For a rational function \(f(x)=P(x)/Q(x)\), suppose \(P\) has multiplicity \(r\) at \(a\) and \(Q\) has multiplicity \(s\) there. Locally, their factorizations give
If \(s>r\), cancellation leaves a denominator factor \((x-a)^{s-r}\). This remaining exponent \(m=s-r\) is the effective pole order. The value of the original expression may be undefined at \(a\), but cancellation is valid for nearby \(x\ne a\), which is exactly where the one-sided limits are studied. If \(s\le r\), no denominator factor remains after cancellation, so this analysis does not give an infinite limit at \(a\).
The One-Sided Sign Rule
Once the function is written in this form, the sign of \(h(x)\) near \(a\) is the same as the sign of \(h(a)\). This follows from continuity: because \(h(a)\ne 0\), values of \(h(x)\) sufficiently close to \(a\) remain on the same side of zero. The sign of \((x-a)^m\) is positive to the right of \(a\). To the left, it is positive for even \(m\) and negative for odd \(m\).
| Pole order | Sign of \(h(a)\) | As \(x\to a^-\) | As \(x\to a^+\) |
|---|---|---|---|
| Even | Positive | \(+\infty\) | \(+\infty\) |
| Even | Negative | \(-\infty\) | \(-\infty\) |
| Odd | Positive | \(-\infty\) | \(+\infty\) |
| Odd | Negative | \(+\infty\) | \(-\infty\) |
The rule describes the direction of the unbounded values, not just their signs. Since \(h(x)\) approaches the nonzero number \(h(a)\), its magnitude stays bounded below by a positive number near \(a\). Meanwhile, \(|x-a|^m\) approaches zero. Therefore the magnitude of \(h(x)/(x-a)^m\) grows without bound.
Deriving the Rule
Here is a direct justification. Let \(f(x)=h(x)/(x-a)^m\), where \(h\) is continuous near \(a\) and \(h(a)\ne 0\). Continuity gives \(h(x)\to h(a)\), so \(h(x)\) has the same sign as \(h(a)\) sufficiently close to \(a\). Also, \(|x-a|^m\to 0\) as \(x\to a\), so the magnitude \(|f(x)|=|h(x)|/|x-a|^m\) tends to infinity.
Since \(h(a)\ne 0\) and \(h\) is continuous, \(h(x)\) keeps the sign of \(h(a)\) sufficiently close to \(a\).
For \(x>a\), \((x-a)^m>0\). For \(x<a\), its sign is positive if \(m\) is even and negative if \(m\) is odd.
Because \(|h(x)|\) stays bounded below by a positive constant while \(|x-a|^m\to 0\), \(|f(x)|\to\infty\) on either side.
Combining these facts proves the sign rule. An even pole order gives the same sign on both sides; an odd pole order gives opposite signs. The sign of \(h(a)\) tells us which direction belongs to each side.
Worked Examples
Worked Example: An Odd Pole with Positive Remaining Factor
Determine the one-sided infinite limits of \(f(x)=\dfrac{x+1}{(x-2)^3(x+4)}\) as \(x\to 2\).
The denominator contains \((x-2)^3\), and the numerator is nonzero at \(2\), since \(2+1=3\). Thus the effective pole order is \(m=3\). Write
Evaluate the remaining factor at the point:
For \(x<2\), \(x-2<0\), so \((x-2)^3<0\). The numerator factor \(h(x)\) remains positive near \(2\), so \(f(x)\) is negative on the left and its magnitude grows without bound. For \(x>2\), \((x-2)^3>0\), so \(f(x)\) is positive and unbounded. Therefore
Check: The factor \(x+4\) is positive near \(2\), as is \(x+1\). The only sign change across \(2\) comes from the odd power \((x-2)^3\), confirming the opposite one-sided signs.
Worked Example: Cancellation Changes the Effective Order
Determine the one-sided limits of \(g(x)=\dfrac{(x-3)^2}{(x-3)^5(x+1)}\) as \(x\to 3\).
The original denominator has multiplicity \(5\) at \(3\), while the numerator cancels two powers. The effective pole order is therefore \(5-2=3\), not \(5\). For \(x\ne 3\), simplify:
The remaining factor has value
The effective order \(3\) is odd. On the left of \(3\), \((x-3)^3<0\); on the right, \((x-3)^3>0\). Since \(h(x)>0\) near \(3\), it follows that
Check: The simplified expression has exactly three uncancelled powers of \(x-3\) in the denominator. The factor \(x+1\) is positive near \(3\), so the sign changes only because that remaining power is odd.
Worked Example: An Even Pole with a Negative Remaining Factor
Determine the one-sided limits of \(r(x)=-\dfrac{x+2}{(x-1)^4(x+3)}\) as \(x\to 1\).
There is no cancellation at \(x=1\), so the pole order is \(m=4\), which is even. Set
At \(x=1\),
An even power \((x-1)^4\) is positive on both sides of \(1\). The numerator factor \(h(x)\) stays negative near \(1\), so \(r(x)\) tends to negative infinity from both sides:
Check: Near \(1\), both \(x+2\) and \(x+3\) are positive, so the leading minus sign makes the numerator factor negative. The fourth power in the denominator is positive on each side. This confirms the two negative signs.
A Common Mistake: Using the Uncancelled Power
A frequent error is to decide whether the one-sided signs agree by looking at the denominator’s original multiplicity, without checking whether the numerator cancels some of those powers. In the second example, treating the original power \(5\) as the effective order would happen to predict opposite signs because \(5\) is odd. But it would still give the wrong pole order and would lead to incorrect conclusions in other cases.
For instance, if the denominator has multiplicity \(4\) and the numerator cancels one power, the effective order is \(4-1=3\), which is odd. Using the original even multiplicity would incorrectly predict the same sign on both sides. Always factor both polynomials and subtract the multiplicities before applying the parity rule.
Check Your Understanding
For each question, identify any cancellation and use the remaining factor’s sign to determine the one-sided behavior.
- For \(f(x)=\dfrac{2}{(x+1)^2}\), do the one-sided limits at \(x=-1\) have the same sign or opposite signs? What are they?
- For \(g(x)=\dfrac{x-2}{(x-2)^4(x+1)}\), find the effective pole order at \(x=2\) and determine both one-sided limits.
- For \(h(x)=-\dfrac{1}{(x-4)^3}\), determine the limits as \(x\to4^-\) and \(x\to4^+\).
- If the numerator and denominator both have a zero of multiplicity \(3\) at \(a\), does their quotient necessarily have an infinite limit there? Explain using cancellation.
- Why does a nonzero value of \(h(a)\) ensure that \(h(x)/(x-a)^m\) is unbounded in magnitude near \(a\)?
