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Removing Discontinuities

Repair a removable discontinuity or choose a piecewise parameter by matching both sides and the target value.

This is shared AB/BC instruction. Open the AB edition of Topic 1.13. Your BC practice and progress stay in this book.

College Board Topic 1.13 · CED p. 45 (PDF p. 50) · CED effective Fall 2020; project course outline 2026–27.

To repair a removable discontinuity, define the target value to equal the finite common limit. This changes one value while keeping the nearby rule. If the one-sided limits disagree or are unbounded, no single finite target value can create continuity.

For f(x)=(x²−4)/(x−2) when x≠2, nearby values equal x+2, so define f(2)=4. For a piecewise boundary, set the left limiting expression equal to the right limiting expression and to the included target value. Solve those equations; do not merely make the formulas resemble each other.

For f(x)=ax+1 when x<2 and f(x)=x²−1 when x≥2, the boundary value is 3. The left limit is 2a+1, so continuity requires 2a+1=3 and a=1. Both pieces are already continuous away from 2.

Let's work through it

  1. Find lim(x→2)(x²−4)/(x−2)=lim(x→2)(x+2)=4. Reason: cancellation preserves the nearby values.
  2. Define f(2)=4. Reason: the continuity definition requires value and limit to agree.
  3. Check both one-sided limits and the new value are 4. Reason: a repair is complete only when all continuity conditions hold.

Try it from memory

Pause & retrieve

Try it before you peek.

Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.

Question 1 of 20 self-assessed

Question 1

What value fills the hole of (x²−9)/(x−3) at 3?

See every recall question with its answer

These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.

Retrieval 1

What value fills the hole of (x²−9)/(x−3) at 3?

Worked answer and reason

6.

The nearby expression is x+3.

Retrieval 2

Can one chosen value repair left limit 1 and right limit 2?

Worked answer and reason

No.

The common finite limit required for a removable discontinuity is missing.

Catch the tempting mistake

Choose any point between the two one-sided limits to repair a jump.
A chosen point cannot change either nearby trace; the two limits must already match.
Calculator workspace and limits

For calculus, set the angle selector to Radians. This scientific calculator evaluates arithmetic and functions. It does not graph arbitrary functions, find numerical roots, evaluate numerical derivatives, or compute definite integrals. Those tasks need an approved graphing tool; the May 2027 exam permits approved handheld graphing calculators and Bluebook’s built-in Desmos on calculator-required parts. A graph or decimal estimate helps you check a conclusion; your setup, hypotheses, units and justification still do the explaining.

Scientific calculator

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Put it to work

For the teacher
  • Have students check their parameter by substituting into both boundary expressions.
  • Include a “no possible repair” problem, so students do not assume every parameter problem has a solution.

Unit planning and formative guidance →

Glossary and deeper reading · 4 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

  • Removable discontinuity

    A removable discontinuity occurs when a finite two-sided limit exists but the function is missing or has a different value at the target. Changing or defining that single value to equal the limit repairs continuity. For example, (x²−1)/(x−1) has a removable discontinuity at 1 and can be repaired by setting its value there to 2. A jump with unequal one-sided limits cannot be repaired this way.

  • Limit

    A limit describes the value approached by outputs as inputs approach a specified target or grow without bound. For an input approaching a finite target, its value at that exact input does not determine the limit. For example, (x²−1)/(x−1) approaches 2 as x approaches 1 even though the original expression is undefined at 1. A finite two-sided limit requires the left and right approaches to agree.

    Read more: 1.2 · Limit →

  • One-sided limit

    A one-sided limit uses inputs approaching a target from just one side. A left-hand limit uses smaller inputs, and a right-hand limit uses larger inputs. For example, a function that equals 0 left of 2 and 1 right of 2 has different one-sided limits at 2. The two-sided limit exists as a finite real number only when both one-sided limits exist and agree.

    Read more: 1.2 · One-sided limit →

  • Continuity

    A function is continuous at an input when its value exists, its finite limit exists, and those two values agree. At a domain endpoint, continuity uses the relevant one-sided approach. For example, x² is continuous at 3 because both its value and its nearby limit are 9. Continuity on an interval requires the appropriate condition at every input in that interval.

    Read more: 1.11 · Continuity →

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