AP CALCULUS BC • LIMITS AND CONTINUITY

Removing Discontinuities

Redefining a function at a single point to restore continuity where a limit already exists.

Historical Context & Motivation

The concept of removing discontinuities is intimately tied to the broader history of rigorizing the notion of continuity and limits in mathematical analysis. For centuries, mathematicians worked with functions intuitively, assuming that algebraic expressions behaved "nicely" everywhere they appeared. It was only when paradoxes and contradictions arose — particularly around infinite series, oscillating functions, and points where formulas broke down — that the need for precise definitions became apparent. The journey from informal manipulation to the modern epsilon-delta framework is one of the great intellectual achievements of nineteenth-century mathematics, and the ability to identify and repair removable discontinuities sits at its heart.

1821
Cauchy's Cours d'Analyse
Augustin-Louis Cauchy published the first rigorous treatment of limits and continuity, defining a function as continuous when "infinitely small increments in the variable produce infinitely small changes in the function." This framework made it possible to classify points where continuity failed.
1861
Weierstrass Formalizes Limits
Karl Weierstrass introduced the ε-δ definition of a limit during his Berlin lectures, replacing Cauchy's intuitive language of "infinitely small" quantities with precise quantifiers. This rigorous foundation allowed mathematicians to distinguish between different types of discontinuities with full logical clarity.
1870s
Classification of Discontinuities
Building on Weierstrass's framework, analysts classified discontinuities into removable, jump, and essential (infinite or oscillatory) types. The key insight emerged: if a two-sided limit exists at a point, the discontinuity is merely an artifact of how the function was defined there and can be "removed" by redefining the function's value.
1893
Jordan's Cours d'Analyse
Camille Jordan's influential textbook systematized the classification of discontinuities and emphasized that removable discontinuities are fundamentally different from jump or essential discontinuities. His treatment solidified the idea that continuity can be restored at isolated points through careful redefinition — a technique central to modern real analysis.

The central question this topic addresses is deceptively simple: if a function fails to be continuous at a single point, but the limit from both sides exists and agrees, can we "fix" the function? The answer is yes — and understanding exactly when and how to perform this repair is an essential skill in calculus. On the AP Calculus BC exam, you will encounter functions defined by algebraic expressions, piecewise rules, or rational forms that have isolated holes, and you must determine the value that restores continuity. This section of the course bridges the gap between evaluating limits and understanding what it truly means for a function to be continuous.

Core Principles & Definitions

Before we can remove a discontinuity, we need to be precise about what continuity requires and how it can fail. Recall that a function f is continuous at a point c if and only if three conditions hold simultaneously: f(c) is defined, the limit of f(x) as x approaches c exists, and that limit equals f(c). A removable discontinuity occurs when the two-sided limit exists but either f(c) is undefined or f(c) does not equal the limit. In contrast, a jump discontinuity arises when the left-hand and right-hand limits both exist but disagree, and an infinite (essential) discontinuity occurs when at least one side diverges to infinity. Only the first type — the removable discontinuity — can be repaired by redefining the function at a single point.

1

Three Conditions for Continuity

At x = c, continuity requires: (1) f(c) is defined, (2) lim(x→c) f(x) exists, and (3) lim(x→c) f(x) = f(c). All three must hold; failure of any one produces a discontinuity.
2

Removable Discontinuity

The two-sided limit L = lim(x→c) f(x) exists, but either f(c) is undefined or f(c) ≠ L. The graph shows a "hole" — the curve approaches a definite height but the point is missing or misplaced.
3

The Repair Strategy

Define (or redefine) f(c) = L. This single-point adjustment makes the function continuous at c because all three continuity conditions are now satisfied. No change is needed anywhere else.
4

Non-Removable Discontinuities

Jump discontinuities (left limit ≠ right limit) and infinite discontinuities (limit diverges) cannot be removed by redefining f at a single point. Recognizing this distinction is critical on the AP exam.
KEY TAKEAWAY
Think of a removable discontinuity like a missing tile in an otherwise perfectly laid mosaic floor. The pattern on either side tells you exactly which tile belongs in the gap. You don't need to redesign the entire floor — you just need to place the correct tile (the limit value) into the empty spot. If the pattern doesn't converge to a single tile shape (as with jump or infinite discontinuities), no single tile can fix the problem.

Visual Explanation

Anatomy of a Removable Discontinuity

The graph of f(x) = (x² − 4)/(x − 2) is identical to the line y = x + 2 everywhere except at x = 2, where the original function is undefined. The open circle at (2, 4) represents the removable discontinuity. Defining f(2) = 4 fills the hole and makes the function continuous.

In the diagram above, observe that the function f(x) = (x² − 4)/(x − 2) can be factored as (x − 2)(x + 2)/(x − 2), which simplifies to x + 2 for all x ≠ 2. The graph is therefore a straight line with slope 1 and y-intercept 2, except at x = 2 where the original expression is undefined (0/0). Both the left-hand limit and the right-hand limit as x → 2 equal 4, so the two-sided limit exists and equals 4. The open circle at (2, 4) visually represents the missing point — the removable discontinuity. By defining f(2) = 4, we fill in the hole and the function becomes the complete, unbroken line y = x + 2. This process of algebraic simplification followed by point-evaluation is the core technique you will use throughout this topic.

Mathematical Framework

The formal machinery behind removing discontinuities rests on the ε-δ definition of limits and the three-part definition of continuity. Here we present the key equations and logical structure that govern the process.

DEFINITION OF CONTINUITY
f is continuous at c ⟺ lim(x→c) f(x) = f(c)
This single equation encodes three requirements: f(c) must be defined (right side meaningful), the limit must exist (left side meaningful), and the two must be equal.
REMOVABLE DISCONTINUITY CONDITION
lim(x→c) f(x) = L exists, but f(c) ≠ L or f(c) is undefined
When the two-sided limit L exists but the function value at c does not match L (or does not exist), the discontinuity is classified as removable. The repair is to set f(c) = L.
ALGEBRAIC SIMPLIFICATION TECHNIQUE
f(x) = g(x) · h(x) / h(x) = g(x) for x ≠ c, where h(c) = 0
When a common factor h(x) causes 0/0 at x = c, cancel it to obtain g(x). Then L = g(c). This technique applies to factorable polynomials, rationalized radical expressions, and trigonometric identities.
PIECEWISE CONTINUITY CONDITION
For f(x) = { g(x), x ≠ c; k, x = c }: continuity at c ⟺ k = lim(x→c) g(x)
When a function is defined piecewise with a special value k at x = c and a formula g(x) elsewhere, continuity at c requires choosing k equal to the limit of g(x) as x → c. This is the most direct form of "removing" a discontinuity.

In practice, computing the limit L often requires one of several algebraic techniques: factoring and canceling common factors, multiplying by a conjugate to rationalize a numerator or denominator, applying trigonometric identities (such as the fundamental limit lim(θ→0) sin θ / θ = 1), or using L'Hôpital's Rule when the expression yields an indeterminate form 0/0 or ∞/∞. Once L is determined, the removed-discontinuity version of f is simply the piecewise function that equals the original expression away from c and equals L at c. It is worth noting that this redefined function and the original agree everywhere except at the single point x = c, so all derivative and integral computations away from c are unaffected.

Classifying Discontinuities

A critical skill on the AP Calculus BC exam is the ability to quickly classify a discontinuity as removable, jump, or infinite. The classification depends entirely on the behavior of the left-hand limit, the right-hand limit, and the function value at the point in question. The following diagram and table provide a systematic decision framework.

This flowchart guides you through classifying any discontinuity. Follow the decision tree from top to bottom: check whether one-sided limits exist, whether they agree, and whether the function value equals the common limit. Only when the two-sided limit exists but f(c) ≠ L (or f(c) is undefined) is the discontinuity removable.
Summary of discontinuity types and removability
TypeLeft & Right Limitsf(c) ConditionRemovable?
RemovableBoth exist and are equal: Lf(c) ≠ L or f(c) undefinedYes — set f(c) = L
JumpBoth exist but L⁻ ≠ L⁺Any value (or undefined)No
InfiniteAt least one side → ±∞Any value (or undefined)No
OscillatoryAt least one side does not exist (oscillates)Any value (or undefined)No

Worked Example

Removing a Discontinuity in a Rational Function

Consider the piecewise function f(x) = (x² − 5x + 6)/(x − 3) for x ≠ 3, and f(3) = k. Determine the value of k that makes f continuous at x = 3.

Finding k for Continuity
1
Step 1 — Identify the discontinuityAt x = 3, the denominator (x − 3) equals zero, and the numerator x² − 5x + 6 = 9 − 15 + 6 = 0. The expression takes the indeterminate form 0/0, which signals a potential removable discontinuity. We need to compute the limit as x → 3.
2
Step 2 — Factor the numeratorFactor x² − 5x + 6 by finding two numbers that multiply to 6 and add to −5. These are −2 and −3, so x² − 5x + 6 = (x − 2)(x − 3). The function becomes f(x) = (x − 2)(x − 3)/(x − 3) for x ≠ 3.
f(x) = (x − 2)(x − 3) / (x − 3)
3
Step 3 — Cancel the common factorSince x ≠ 3, the factor (x − 3) cancels from numerator and denominator, leaving the simplified form f(x) = x − 2 for all x ≠ 3. This algebraic simplification is valid everywhere except at the point of discontinuity.
f(x) = x − 2, x ≠ 3
4
Step 4 — Evaluate the limitSubstitute x = 3 into the simplified expression: lim(x→3) f(x) = lim(x→3) (x − 2) = 3 − 2 = 1. Both one-sided limits agree, confirming that the two-sided limit exists and equals 1.
lim(x→3) f(x) = 1
5
Step 5 — Set f(3) equal to the limitFor continuity at x = 3, we need f(3) = lim(x→3) f(x). Therefore k = 1. The completed continuous function is f(x) = x − 2 for all real numbers.
k = 1
💡 AP Exam Tip
On free-response questions, always state explicitly that the limit exists, show the algebraic simplification, and conclude by verifying that the redefined value satisfies all three continuity conditions. Graders award points for clear reasoning at each stage, not just the final numerical answer.

Techniques & Common Pitfalls

Removing discontinuities is not always as straightforward as factoring a quadratic. Many AP problems involve radical expressions, trigonometric functions, or exponential-logarithmic combinations that require different algebraic strategies. The table below summarizes the most common techniques and the contexts in which they appear.

Algebraic techniques for evaluating limits at removable discontinuities
TechniqueWhen to UseExample
Factor & CancelPolynomial numerator and denominator share a common root(x² − 9)/(x − 3) → x + 3
Conjugate MultiplicationRadical expression yields 0/0; multiply by conjugate to rationalize(√(x+1) − 2)/(x − 3) → 1/(√(x+1) + 2)
Trig Identitiessin θ/θ type expressions or Pythagorean identities create 0/0(1 − cos x)/x² → 1/2 via half-angle
L'Hôpital's Rule0/0 or ∞/∞ form where algebraic techniques are cumbersomelim(x→0) (eˣ − 1)/x = eˣ/1 |₍ₓ₌₀₎ = 1
COMMON PITFALL
Students often confuse the location of a vertical asymptote (infinite discontinuity) with a hole (removable discontinuity). After factoring a rational function, a canceled factor corresponds to a hole, while a factor that remains in the denominator corresponds to a vertical asymptote. For instance, in f(x) = (x − 1)/((x − 1)(x + 2)), the factor (x − 1) cancels, producing a hole at x = 1, but the remaining factor (x + 2) in the denominator produces a vertical asymptote at x = −2. Mistaking one for the other is one of the most frequent errors on the AP exam.

Connections to Advanced Topics

Removing discontinuities is not merely an isolated limit-computation technique — it connects to several advanced topics that appear later in the AP Calculus BC curriculum and beyond. Understanding these connections deepens your appreciation for why continuity matters and how removable discontinuities appear in broader mathematical contexts.

How removing discontinuities connects to advanced calculus and analysis
This TopicAdvanced Connection
Removing a discontinuity by redefining f(c)Definition of the derivative: f′(a) = lim(h→0) [f(a+h) − f(a)]/h is itself the process of evaluating (and removing) a 0/0 discontinuity in the difference quotient
Factoring and canceling common factorsPartial fraction decomposition for integration — identifying and canceling factors is a prerequisite skill for decomposing rational integrands
Piecewise continuity at a pointImproper integrals: when an integrand has a removable singularity inside the interval, redefining the function there allows the Riemann integral to exist without splitting into improper form
Recognizing indeterminate forms 0/0Taylor and Maclaurin series: removable discontinuities in expressions like sin(x)/x are resolved by recognizing the series expansion — the function has a natural continuous extension
Continuous extensions of functionsIn real analysis and topology, the concept generalizes to extending functions from dense subsets to complete spaces — the foundation of measure theory and functional analysis

Perhaps the most important connection for the AP BC exam is to the definition of the derivative itself. When you compute f′(a) = lim(h→0) [f(a + h) − f(a)]/h, the difference quotient is undefined at h = 0 (it yields 0/0). The derivative exists precisely when this removable discontinuity can be "filled in" — that is, when the limit exists. In this sense, every derivative computation is an exercise in removing a discontinuity from the difference quotient. This perspective also clarifies why differentiability implies continuity: if the 0/0 in the difference quotient can be resolved, the original function must have been continuous at that point.

Practice Problems

1
The function g is defined for all real numbers. The graph of g has a hole at the point (3, 5). Which of the following must be true?
2
Let f(x) = (x² − 16)/(x − 4). What value should be assigned to f(4) to make f continuous at x = 4?
3
The function h is defined by h(x) = (√(x + 5) − 3)/(x − 4) for x ≠ 4, and h(4) = k. Find the value of k that makes h continuous at x = 4.
PROBLEM 4APPLIED
A particle moves along the x-axis, and its position is given by s(t) = (t³ − 8)/(t − 2) for t ≠ 2 (time in seconds, position in meters). The position sensor fails at t = 2. (a) Find the position the particle would occupy at t = 2 if the motion were continuous. (b) Find the velocity of the particle at t = 2. Justify that this velocity exists. (c) Explain why the position function has a removable discontinuity at t = 2 rather than a jump or infinite discontinuity.
PROBLEM 5CRITICAL THINKING
Let f(x) = (sin(πx))/(x² − 1). (a) Identify all points where f has discontinuities and classify each as removable or non-removable. (b) For each removable discontinuity, find the value that would make f continuous at that point. (c) Justify your classification using limits.

Summary

A removable discontinuity occurs at x = c when the two-sided limit L = lim(x→c) f(x) exists but either f(c) is undefined or f(c) ≠ L. The discontinuity is "removed" by defining or redefining f(c) = L, which satisfies all three conditions for continuity at that point. In contrast, jump discontinuities (one-sided limits disagree) and infinite discontinuities (limits diverge) cannot be repaired by a single-point redefinition.

To find L, use algebraic techniques such as factoring and canceling, conjugate multiplication, trigonometric identities, or L'Hôpital's Rule to resolve the 0/0 indeterminate form. This skill is foundational: every derivative computation involves removing a 0/0 discontinuity from the difference quotient, making the concept of removable discontinuities central to all of calculus.

Varsity Tutors • AP Calculus BC • Removing Discontinuities