Why Simplification Comes Before Classification
A rational expression can look singular because its denominator is zero at a point. But the expression as written may also have a numerator that vanishes there. Simplifying common factors reveals whether the apparent problem is a removable hole or whether a denominator factor remains and the function is still unbounded.
The order matters. First record where the original expression is defined; then simplify and examine the remaining factors. Cancellation can describe the function near a hole, but it does not change the original domain or automatically assign a value at the missing point.
Rational Functions and Their Original Domains
Let \(P\) and \(Q\) be polynomials, with \(Q\) not identically zero. The rational function \(R(x)=P(x)/Q(x)\) is defined at precisely those real numbers \(x\) for which \(Q(x)\ne 0\). A zero \(a\) of \(Q\) is therefore excluded from the original domain, even if a factor in the numerator can later be canceled.
For polynomials, the classification can be made by comparing the multiplicities of the factor \(x-a\). The multiplicity is the number of times that factor occurs in a polynomial’s factorization. The zero polynomial requires separate treatment: it does not have a finite exact multiplicity of \(x-a\).
- If \(m\ge n\), the singularity is removable. Its extension value is \(p(a)/q(a)\) when \(m=n\), and \(0\) when \(m>n\).
- If \(m<n\), the quotient has a pole of order \(n-m\) at \(a\).
Why the Multiplicity Test Works
For \(x\ne a\), canceling the displayed common factors gives the identity
Because \(q(a)\ne 0\), the function \(p(x)/q(x)\) is continuous near \(a\), and its limit there is the nonzero number \(p(a)/q(a)\). The factor \((x-a)^{m-n}\) determines what happens to the product.
If \(m=n\), the power of \(x-a\) is zero, so the limit is \(p(a)/q(a)\), a finite nonzero number.
If \(m>n\), the remaining positive power of \(x-a\) tends to zero, so the limit is zero.
If \(m<n\), the remaining factor is \(1/(x-a)^{n-m}\). The other factor tends to a nonzero number, so the quotient is unbounded.
For completeness, in the last case let \(k=n-m>0\) and \(c=p(a)/q(a)\ne 0\). By continuity, for \(x\) sufficiently close to \(a\), \(\left|p(x)/q(x)\right|>|c|/2\). Therefore
The right side tends to infinity as \(x\to a\), proving that the quotient is unbounded. If \(P\equiv0\), no multiplicity \(m\) is used: the quotient is simply zero wherever it is defined, so its limit at \(a\) is zero.
Worked Example: Equal Multiplicities Give a Removable Hole
Worked Example: Equal Multiplicities Give a Removable Hole
Consider \(R(x)=(x^2-1)/(x-1)\), originally defined for \(x\ne1\). Factor the numerator:
At \(a=1\), the numerator and denominator each have multiplicity one, so \(m=n=1\). The limit and extension value are
The original expression is still undefined at \(x=1\); the simplified expression describes its values only for \(x\ne1\). If we evaluate its integral across the hole, split at \(1\) and take the one-sided limits:
Thus the integral across the removable hole is \(3/2+5/2=4\). Check: the extension \(x+1\) has antiderivative \(x^2/2+x\), whose derivative is \(x+1\); evaluating from \(0\) to \(2\) gives \(4-0=4\), agreeing with the split calculation.
Worked Example: Extra Numerator Factors Still Leave a Hole
Worked Example: Extra Numerator Factors Still Leave a Hole
Consider \(S(x)=(x-2)^3(x+1)/(x-2)^2\), defined originally for \(x\ne2\). At \(a=2\), the multiplicities are \(m=3\) and \(n=2\). Cancel only the two common factors:
One factor \(x-2\) remains in the numerator, so the limit at the hole is \(0\). To evaluate the integral from \(1\) to \(3\), split at \(2\) and use the simplified expression on each punctured interval. An antiderivative is
The left-hand contribution is
The right-hand contribution is
The total is \(-7/6+11/6=2/3\). Check by evaluating the extension over the whole interval: \(F(3)-F(1)=-3/2-(-13/6)=2/3\). Also, differentiating \(F\) gives \(x^2-x-2=(x-2)(x+1)\), as required.
Worked Example: A Residual Denominator Factor Is a Pole
Worked Example: A Residual Denominator Factor Is a Pole
Consider \(T(x)=(x-1)/(x-1)^3\), defined for \(x\ne1\). The numerator has multiplicity \(m=1\), while the denominator has multiplicity \(n=3\). Cancel one common factor, not all three:
The remaining denominator factor gives a pole of order \(2\). On each side of \(1\), the integrand is positive. For \(0\le x<1\), an antiderivative is \(-1/(x-1)\), so
For \(1<x\le2\), the corresponding truncated integral is
Both one-sided integrals diverge, so the integral across \(1\) does not converge. Check: differentiating \(-1/(x-1)\) gives \(1/(x-1)^2\), confirming the antiderivative and the displayed truncated evaluations.
Cancellation Does Not Change the Original Domain
The equality obtained by canceling a factor is valid only at points where the original quotient is defined. For example, \((x^2-1)/(x-1)=x+1\) holds for \(x\ne1\), not at \(x=1\). The simplified formula supplies the limit there, but assigning that value creates an extension rather than changing the original formula’s domain.
For an integral, a removable hole can be handled by taking the one-sided limits at that point. When the extension is continuous, those limits are finite and agree with integrating the continuous extension. By contrast, a genuine pole must be treated as an improper singularity: the integral across an interior point converges only if both one-sided integrals converge.
A Reliable Classification Procedure
At a suspected denominator zero, use the following sequence. It keeps the original domain separate from the behavior of the simplified expression.
Check that \(Q(a)=0\); the original quotient is undefined at \(a\).
If \(P\not\equiv0\), write the powers of \(x-a\) in numerator and denominator. If \(P\equiv0\), handle the quotient separately as zero on its domain.
Equal or greater numerator multiplicity gives a removable hole; greater denominator multiplicity gives a pole.
For a removable hole, use the finite extension when evaluating the integral. For a pole, test the required one-sided improper integrals.
The key point is that algebraic simplification is not just a cosmetic step. It reveals the actual local behavior of the quotient, while preserving the distinction between the original function and its continuous extension. Correct classification prevents a harmless hole from being mistaken for a pole—and prevents a residual pole from being overlooked.
Check Your Understanding
For each question, keep the original domain in view as well as the simplified expression.
- For \((x^2-9)/(x-3)\), what is the original excluded point, and what is the extension value there?
- If the numerator has multiplicity \(4\) at \(a\) and the denominator has multiplicity \(2\), is the singularity removable or a pole? What is the limit?
- If the numerator has multiplicity \(1\) and the denominator has multiplicity \(3\), what is the order of the pole?
- How should the case \(P\equiv0\) be classified when \(Q(a)=0\)?
- Why does canceling a common factor not make the original rational expression defined at the canceled zero?
