A removable discontinuity (a hole) occurs in a rational function at x = c when the multiplicity of the real zero c in the numerator is greater than or equal to its multiplicity in the denominator; the function is undefined at c, but output values near c approach a single value L, putting the hole at (c, L).
A removable discontinuity is the formal name for a hole in the graph of a rational function. It happens when the numerator and denominator share a common factor. The function is undefined at that input value (because the denominator equals zero there), but unlike a vertical asymptote, the graph doesn't blow up to infinity. It just skips one single point.
The CED gives you a clean test (1.10.A.1). Look at the multiplicity of the real zero in the numerator versus the denominator. If the numerator's multiplicity is greater than or equal to the denominator's, you get a hole at that x-value. To find the hole's exact location (1.10.A.2), examine output values for inputs close to c. If those outputs get arbitrarily close to some value L, the hole sits at the point (c, L). In practice, you cancel the common factor and plug c into the simplified function. The reason it's called "removable" is that you could patch the function by defining its value at that one point to be L, and the discontinuity disappears.
Removable discontinuities live in Topic 1.10 (Rational Functions and Holes) in Unit 1: Polynomial and Rational Functions, supporting learning objective 1.10.A: determine holes in graphs of rational functions. This is one of the four behaviors a rational function can show at a problem input (hole, vertical asymptote, plus end behavior features like horizontal or slant asymptotes), and the exam expects you to tell them apart fast. The multiplicity comparison is the deciding rule, so this term ties directly back to your polynomial zero work earlier in Unit 1. It's also your first real taste of limit thinking: "outputs get arbitrarily close to L as inputs get close to c" is exactly the idea AP Calculus formalizes later.
Keep studying AP® Precalculus Unit 1
Multiplicity (Unit 1)
Multiplicity is the entire decision rule for holes. If a real zero appears in the numerator with multiplicity greater than or equal to its multiplicity in the denominator, the graph has a hole there. If the denominator wins, you get a vertical asymptote instead. One comparison, two totally different graph features.
Real Zeros of Polynomials (Unit 1)
You can't find a hole without first factoring. Every hole comes from a real zero shared by the numerator and denominator, so the factoring skills from earlier in Unit 1 are the setup work for every Topic 1.10 problem.
Vertical Asymptotes of Rational Functions (Unit 1)
Holes and vertical asymptotes both come from zeros of the denominator, which is why they get confused. The difference is whether the factor cancels. A canceled factor leaves a hole; a leftover denominator factor sends outputs to positive or negative infinity.
Limits in AP Calculus (looking ahead)
The CED's hole-location idea, that inputs sufficiently close to c give outputs arbitrarily close to L, is the informal definition of a limit. When you compute the y-coordinate of a hole, you're computing the limit of the function as x approaches c. AP Calc just gives it notation.
This is tested as a multiple-choice skill with a few standard angles. You might get a function like h(x) = (x² − 4x + 3)/(x² − 3x + 2) and be asked to find the holes. Factor both, spot the shared (x − 1), cancel, and evaluate the simplified function at x = 1 to get the hole at (1, 2). Questions also run in reverse: given that a function has a hole at x = 3 with y-coordinate 4, pick the rational function that produces it, or, given a simplified form like g(x) = x² − 5x + 6 valid for all x ≠ 2, reconstruct an original function with a factor of (x − 2) on top and bottom. The most common trap is conceptual. If a factor like (x + 4) appears once in the numerator and once in the denominator, the answer is a hole at x = −4, not a vertical asymptote. No released FRQ has used the phrase "removable discontinuity" verbatim, but analyzing rational function behavior at undefined inputs is core Unit 1 territory.
Both happen where the denominator equals zero, but they look and behave completely differently. A removable discontinuity is a single missing point; outputs near it stay close to a finite value L. A vertical asymptote is a non-removable discontinuity; outputs explode toward positive or negative infinity. The multiplicity comparison settles it. Numerator multiplicity ≥ denominator multiplicity means hole. Denominator multiplicity greater means vertical asymptote. Quick check on h(x) = (x − 1)(x − 3) / [(x − 1)(x − 2)]: x = 1 cancels, so it's a hole; x = 2 doesn't, so it's an asymptote.
A removable discontinuity is a hole in a rational function's graph, occurring at an input where the function is undefined but the outputs nearby approach one finite value.
Per EK 1.10.A.1, a hole occurs at x = c when the multiplicity of c as a zero of the numerator is greater than or equal to its multiplicity in the denominator.
To find the y-coordinate of a hole, cancel the common factor and evaluate the simplified function at x = c; the hole is the point (c, L).
If the denominator's multiplicity at a zero is greater than the numerator's, you get a vertical asymptote there, not a hole.
A common factor appearing once in the numerator and once in the denominator, like (x + 4), creates a hole, never an asymptote.
The idea that inputs close to c give outputs arbitrarily close to L is exactly the limit concept you'll formalize in AP Calculus.
It's a hole in the graph of a rational function at an input x = c where the function is undefined, but nearby outputs approach a single value L. It comes from a common factor in the numerator and denominator, and the hole sits at the point (c, L).
No. Both occur at zeros of the denominator, but a hole is one missing point where nearby outputs stay near a finite value, while a vertical asymptote sends outputs to positive or negative infinity. The multiplicity comparison from EK 1.10.A.1 tells you which one you have.
Cancel the common factor, then plug the hole's x-value into the simplified function. For h(x) = (x² − 4x + 3)/(x² − 3x + 2), the factor (x − 1) cancels, and the simplified function (x − 3)/(x − 2) gives 2 at x = 1, so the hole is at (1, 2).
No. The simplified function equals the original everywhere except at x = c, where the original is still undefined. That's why answers like g(x) = x² − 5x + 6 come with the restriction "for all x ≠ 2." The hole is part of the original function's graph.
You get a hole. EK 1.10.A.1 says a hole occurs when the numerator's multiplicity is greater than OR equal to the denominator's. So a factor like (x + 4) appearing once on top and once on bottom creates a hole at x = −4.
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