Why Factor Before Classifying
A rational expression is a quotient of polynomials. Its denominator determines which real inputs are excluded from its domain. But an excluded input does not always produce the same kind of discontinuity: the function may approach a finite value, or its values may become unbounded near that input.
Factoring the numerator and denominator makes the distinction visible. A shared factor may cancel when the input is not its zero, producing a simpler formula that reveals the function’s behavior nearby. The original expression, however, remains undefined at any input that made its original denominator zero. Cancellation does not put that input back into the original domain.
Domain and Discontinuity Types
For a rational expression \(f(x)=P(x)/Q(x)\), where \(P\) and \(Q\) are polynomials and \(Q\) is not the zero polynomial, the domain consists of all real \(x\) for which \(Q(x)\ne 0\). Thus, the real zeros of the original denominator are precisely the excluded inputs to check.
For rational expressions, these types can be distinguished by counting how many copies of the factor \(x-a\) occur in the numerator and denominator. A factor’s multiplicity is the number of times it occurs. For example, \((x-a)^3\) has multiplicity \(3\) at \(a\).
The Multiplicity Test
Suppose the denominator vanishes at \(a\). Factor the numerator and denominator near \(a\) in the form
where \(p(a)\ne 0\), \(q(a)\ne 0\), and \(n\ge 1\). The integers \(m,n\ge 0\) count the multiplicities of the factor \(x-a\) in the numerator and denominator. For \(x\ne a\), cancel the shared factors:
This equality applies near \(a\) only at inputs where the original expression is defined. In particular, it does not define the original function at \(a\).
More precisely, if \(m=n\), the limit is \(p(a)/q(a)\). If \(m>n\), the remaining numerator factor forces the limit to be \(0\). If \(m<n\), the function contains a nonzero factor times \(1/(x-a)^{n-m}\), so its absolute value grows without bound as \(x\) approaches \(a\). The parity of \(n-m\), together with the sign of \(p(a)/q(a)\), determines the one-sided signs.
Why the Multiplicity Test Works
Here is the justification. Since \(p\) and \(q\) are polynomials, they are continuous, and \(q(a)\ne 0\). Therefore, near \(a\), the ratio \(p(x)/q(x)\) approaches the finite nonzero number \(p(a)/q(a)\). The only remaining behavior comes from the power of \(x-a\).
For \(x\ne a\), divide numerator and denominator by the smaller of the two powers of \(x-a\). This gives \(\frac{p(x)}{q(x)}(x-a)^{m-n}\).
Because \(p(a)\ne 0\) and \(q(a)\ne 0\), the ratio \(p(x)/q(x)\) approaches the finite nonzero value \(p(a)/q(a)\).
If \(m-n\ge 0\), the power has a finite limit, so the whole expression does too. If \(m-n<0\), it is a reciprocal positive power of \(x-a\), whose absolute value grows without bound.
This also explains why cancellation alone is not enough to declare a hole. If the denominator has more copies of \(x-a\) than the numerator, cancellation reduces the denominator’s multiplicity but does not eliminate it. The function is still unbounded near \(a\).
Worked Examples
Worked Example: One Removable Input and One Infinite Input
Classify the discontinuities of
First factor the numerator and denominator:
The original denominator is zero at \(x=1\) and \(x=2\), so both inputs are excluded. For \(x\ne 1,2\), cancel the shared factor:
At \(x=1\), the remaining denominator is nonzero, and the limit is
Thus \(x=1\) is removable; assigning the value \(-2\) would fill the hole. At \(x=2\), the simplified numerator approaches \(3\), while the denominator approaches zero. For \(x<2\) close to \(2\), \(x+1>0\) and \(x-2<0\), so \(f(x)\to-\infty\). For \(x>2\) close to \(2\), both are positive, so \(f(x)\to+\infty\). Thus \(x=2\) is an infinite discontinuity.
Check: the simplified formula agrees with the original wherever the original is defined, and its denominator still vanishes at \(2\) but not at \(1\). The two classifications match those facts.
Worked Example: Partial Cancellation Still Leaves an Infinite Discontinuity
Classify the excluded input of
The original denominator vanishes at \(x=3\) and \(x=1\). Focus first on \(x=3\). The numerator has multiplicity \(2\) there, while the denominator has multiplicity \(3\). Cancel two copies of \(x-3\), valid for \(x\ne 3\):
One factor \(x-3\) remains in the denominator. Near \(3\), \(x+2\) is positive and \(x-1\) is positive. The remaining factor \(x-3\) is negative to the left and positive to the right. Therefore,
So \(x=3\) is an infinite discontinuity, not a removable one. Check: the numerator had two copies of the vanishing factor and the denominator had three; after cancellation, exactly one denominator copy remains, consistent with the one-sided behavior.
Worked Example: Classifying Every Excluded Input
Consider
The original denominator is zero at \(x=2\) and \(x=-3\); both are excluded. For \(x\ne 2,-3\), cancel the shared squared factor:
At \(x=2\), the simplified denominator is nonzero, so
Thus \(x=2\) is removable. At \(x=-3\), the simplified numerator approaches \(-2\) and the denominator approaches zero. For \(x<-3\) close to \(-3\), both \(x+1\) and \(x+3\) are negative, so their quotient is positive and grows without bound. For \(x>-3\) close to \(-3\), \(x+1<0\) and \(x+3>0\), so the quotient is negative and unbounded. Hence
Check: cancellation removes the factor causing the exclusion at \(2\), giving the finite limit \(3/5\); no factor \(x+3\) cancels, so the expression remains unbounded near \(-3\).
A Common Mistake: Restoring an Excluded Input
A frequent error is to cancel a factor and then treat the simplified formula as if it were the original function on its entire domain. For example, after simplifying \((x-1)(x+1)/((x-1)(x-2))\) to \((x+1)/(x-2)\), one might claim that the original expression is defined at \(x=1\), because the simplified denominator is not zero there. That claim is wrong: the original denominator \((x-1)(x-2)\) equals zero at \(1\).
The formulas agree at every input in the original domain, but their domains differ. At \(x=1\), the original function is undefined and has a removable discontinuity; the simplified formula gives the value \(2/(-1)=-2\), which is the value that would fill the hole. Keep the original domain when classifying discontinuities.
Why This Classification Is Useful
Factoring provides a systematic way to read local behavior from an algebraic expression. It identifies every excluded input, tells whether cancellation removes all vanishing denominator factors, and supplies the finite limit when a hole is removable. When a factor remains in the denominator, one-sided signs further describe whether values rise or fall without bound.
For each real zero \(a\) of the original denominator, use this checklist:
- Record \(a\) as excluded from the original domain.
- Factor the numerator and denominator and count the multiplicities of \(x-a\).
- If the numerator multiplicity is at least as large, cancel and evaluate the finite limit.
- If the denominator multiplicity is larger, identify the remaining denominator power and, when useful, check one-sided signs.
- State the classification without changing the original domain.
Check Your Understanding
For each question, identify the original excluded inputs before using cancellation.
- For \(\frac{x^2-4}{x^2-x-2}\), factor the expression and classify each excluded input.
- For \(\frac{(x+1)^3}{(x+1)^2(x-2)}\), classify the excluded input \(x=-1\). Is the discontinuity removable or infinite?
- For \(\frac{(x-4)^2}{(x-4)^5}\), how many factors of \(x-4\) remain in the denominator after cancellation, and what type of discontinuity occurs at \(4\)?
- In a rational expression, the numerator and denominator both contain \((x-a)^2\), and the denominator has one additional factor of \(x-a\). What does the multiplicity test imply at \(a\)?
- Why does canceling a factor not, by itself, add its zero to the domain of the original expression?
