Removing Discontinuities - AP Calculus AB

SECTION 1

Historical Context & Motivation

The concept of continuity—and the problems that arise when it fails—has driven some of the most important developments in the history of calculus. Long before rigorous definitions existed, mathematicians recognized that certain expressions break down at particular points even though the overall behavior of the function seems perfectly well-behaved nearby. The desire to "repair" these breakdowns motivated centuries of work on limits, function definition, and the very foundations of analysis. Understanding how and when a removable discontinuity can be eliminated by a simple redefinition reveals a deep principle: the value a function approaches matters more than the value it happens to take.

1684 • Leibniz's Calculus Published

Gottfried Wilhelm Leibniz publishes his differential calculus, operating with an intuitive notion of continuity and implicitly assuming that useful functions are "smooth" wherever they are defined.

1821 • Cauchy's Cours d'analyse

Augustin-Louis Cauchy provides the first rigorous definition of a limit and uses it to define continuity, establishing the framework needed to classify different types of discontinuities.

1861 • Weierstrass Formalizes ε–δ

Karl Weierstrass delivers his ε–δ definition of limits and continuity, giving mathematicians a precise criterion for determining whether a discontinuity is removable or essential.

1900s • Modern Analysis & AP Curriculum

The classification of discontinuities into removable, jump, and infinite types becomes standard in university courses and eventually in AP Calculus, where students learn to identify and remove point discontinuities algebraically.

The central question this lesson addresses is deceptively simple: if a function is undefined or incorrectly defined at a single point, yet the limit exists there, can we patch the hole? The answer is yes—and the technique for doing so is both elegant and heavily tested on the AP Calculus AB exam. Mastering it requires fluency with limits, factoring, and the formal definition of continuity, all of which converge in this single topic.

SECTION 2

Core Principles & Definitions

Before we can remove a discontinuity, we need a precise vocabulary for what continuity means and how it can fail. Recall that a function f is continuous at x = a if and only if three conditions hold simultaneously: f(a) is defined, lim as x→a of f(x) exists, and lim as x→a of f(x) equals f(a). A removable discontinuity arises when the second condition is met—the limit exists—but either the first or third condition fails. This is the only type of discontinuity that can be "removed" by redefining the function at a single point.

Continuity at a Point

A function f is continuous at x = a when f(a) is defined, lim(x→a) f(x) exists, and lim(x→a) f(x) = f(a). All three conditions must hold.

Removable Discontinuity

The limit L = lim(x→a) f(x) exists and is finite, but f(a) ≠ L or f(a) is undefined. A "hole" appears on the graph at (a, L).

Non-Removable Discontinuity

The limit does not exist (jump or infinite discontinuity). No single-point redefinition can restore continuity; the break is structural.

Removing the Discontinuity

Define or redefine f(a) = L. The patched function agrees with the original everywhere except possibly at x = a and is now continuous there.

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KEY TAKEAWAY

Think of a removable discontinuity like a single missing tile in an otherwise perfect mosaic floor. The surrounding pattern tells you exactly which tile belongs there. Removing the discontinuity is simply placing the correct tile—redefining f(a) to equal the limit. A jump or infinite discontinuity is more like a crack running through the floor; no single tile can fix it.

SECTION 3

Visual Explanation

A graph is the most immediate way to see the difference between a removable and a non-removable discontinuity. The diagram below shows a function with a removable discontinuity (a hole) at x = 2 alongside the repaired version of the function. Notice that the curve is identical on both sides of x = 2; only the single point is missing in the original version. The open circle indicates the hole, and the filled circle shows the value that restores continuity.

The open circle at (2, 4) represents the removable discontinuity—the limit exists and equals 4, but the original function either is undefined or takes a different value there (shown in pink at f(2) = 3). Redefining f(2) = 4 fills the hole and makes the function continuous.

The key geometric insight is that the curve passes through the point (2, 4) in a limiting sense—both the left-hand and right-hand limits converge there—but the function either skips or misassigns that y-value. The algebraic simplification (x² − 4)/(x − 2) = (x + 2)(x − 2)/(x − 2) = x + 2 for x ≠ 2 reveals the continuous function hiding behind the fraction. The cancellation of the common factor is the algebraic act of "removing" the discontinuity.

SECTION 4

Mathematical Framework

The formal procedure for removing a discontinuity rests on the definition of continuity and the evaluation of limits. The process has three stages: identify the point of discontinuity, compute the limit, and redefine the function. Each stage connects to specific algebraic techniques that appear repeatedly on the AP exam.

DEFINITION OF CONTINUITY

f is continuous at x = a ⟺ lim(x→a) f(x) = f(a)

This single equation encodes all three conditions: f(a) must exist, the limit must exist, and they must be equal. A removable discontinuity satisfies only the second condition.

REMOVABLE DISCONTINUITY CONDITION

lim(x→a) f(x) = L exists and is finite, but f(a) ≠ L or f(a) is undefined

L is the value the function approaches. If we define g(x) = f(x) for x ≠ a and g(a) = L, then g is continuous at a.

COMMON ALGEBRAIC TECHNIQUE — FACTORING

f(x) = P(x)/Q(x), Q(a) = 0 and P(a) = 0 ⟹ factor and cancel (x − a)

When both numerator and denominator vanish at x = a, the rational expression has a common factor of (x − a). Cancelling yields a simplified expression whose value at a equals the limit of the original.

ALTERNATE TECHNIQUE — RATIONALIZATION

lim(x→a) [√(x + h) − √a] / (x − a) → multiply by [√(x + h) + √a] / [√(x + h) + √a]

When a radical appears, multiplying by the conjugate eliminates the 0/0 indeterminate form and reveals a cancellable factor, allowing direct evaluation of the limit.

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AP Exam Tip

The AP Calculus AB exam frequently presents piecewise-defined functions and asks you to find the value of a constant that makes the function continuous. This is precisely the act of removing a discontinuity: set the function value equal to the limit and solve for the unknown parameter.

SECTION 5

Classifying Discontinuities

Not every discontinuity can be removed, so a critical skill on the AP exam is classifying the type of discontinuity present. The three principal types—removable, jump, and infinite—are distinguished by the behavior of the one-sided limits. The diagram below places all three side by side for comparison, making the visual signatures unmistakable.

Left: a removable discontinuity with an open circle (hole) where both sides approach the same y-value. Center: a jump discontinuity where left-hand and right-hand limits disagree. Right: an infinite discontinuity (vertical asymptote) where at least one side diverges to ±∞.

Property	Removable	Jump	Infinite
lim(x→a⁻) f(x)	= L (finite)	= L₁ (finite)	= ±∞
lim(x→a⁺) f(x)	= L (same)	= L₂ ≠ L₁	= ±∞
Two-sided limit	Exists (= L)	Does not exist	Does not exist
Can be removed?	Yes	No	No

Classification of discontinuities by one-sided limit behavior

The essential diagnostic is straightforward: compute the two-sided limit. If it exists as a finite number L, the discontinuity is removable, and you redefine f(a) = L. If the one-sided limits exist but differ, you have a jump. If either one-sided limit is infinite, you have a vertical asymptote. On the AP exam, the algebraic signal for a removable discontinuity in a rational function is a 0/0 indeterminate form at the point of interest, meaning both numerator and denominator share a common factor.

SECTION 6

Worked Example

Consider the function f(x) = (x² − 5x + 6)/(x² − 4). We want to determine whether f has a removable discontinuity at x = 2 and, if so, redefine the function to make it continuous there.

Removing a Discontinuity in a Rational Function

1
Step 1 — Identify the point of discontinuity

The denominator x² − 4 = (x − 2)(x + 2) equals zero when x = 2 or x = −2. So f is undefined at x = 2, which is a potential discontinuity. Check the numerator: x² − 5x + 6 = (x − 2)(x − 3). At x = 2, the numerator is also 0, giving the indeterminate form 0/0. This signals a possible removable discontinuity.

0/0 indeterminate form at x = 2 → candidate for removable discontinuity
2
Step 2 — Factor and cancel the common factor

Factor completely: f(x) = (x − 2)(x − 3) / [(x − 2)(x + 2)]. For x ≠ 2, cancel the common factor (x − 2) to obtain the simplified function g(x) = (x − 3)/(x + 2).

f(x) = (x − 3)/(x + 2) for x ≠ 2
3
Step 3 — Evaluate the limit

Now compute lim(x→2) f(x) using the simplified form: lim(x→2) (x − 3)/(x + 2) = (2 − 3)/(2 + 2) = (−1)/4 = −1/4. The limit exists and is finite, confirming the discontinuity at x = 2 is removable.

lim(x→2) f(x) = −1/4
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Step 4 — Redefine the function

Define the new function: g(x) = f(x) when x ≠ 2, and g(2) = −1/4. Equivalently, g(x) = (x − 3)/(x + 2) for all x ≠ −2. The function g is now continuous at x = 2 because g(2) = lim(x→2) g(x) = −1/4.

g(2) = −1/4 removes the discontinuity
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Step 5 — Verify and note remaining discontinuity

At x = −2, the denominator of the simplified form (x + 2) is zero while the numerator (x − 3) = −5 ≠ 0. This produces a non-zero over zero form, indicating a vertical asymptote—an infinite discontinuity that cannot be removed. Thus f has two discontinuities: a removable one at x = 2 and a non-removable one at x = −2.

x = −2: infinite (non-removable) discontinuity

SECTION 7

Techniques for Finding Limits at Removable Discontinuities

The 0/0 indeterminate form that signals a removable discontinuity can be resolved by several algebraic techniques. The table below compares the major methods, each suited to different function types. Being fluent in all of them gives you flexibility on the AP exam, where the function may involve polynomials, radicals, or trigonometric expressions.

Technique	When to Use	Example
Factoring	Rational functions where numerator and denominator share a common polynomial factor	(x² − 9)/(x − 3) = (x + 3) for x ≠ 3
Rationalization	Expressions involving square roots that produce 0/0; multiply by the conjugate	(√x − 2)/(x − 4): multiply by (√x + 2)/(√x + 2)
Trig identities	Trigonometric 0/0 forms; use sin(x)/x → 1 or Pythagorean identities	sin(3x)/x = 3 · [sin(3x)/(3x)] → 3 as x → 0
Simplifying complex fractions	Nested fractions that produce 0/0; combine into a single fraction first	(1/x − 1/3)/(x − 3): combine numerator, then cancel
Piecewise redefinition	When f is defined piecewise and a parameter must be chosen to force continuity	f(x) = {x² + k, x ≤ 1; 3x, x > 1}: set 1 + k = 3

Comparison of algebraic techniques for resolving removable discontinuities

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KEY TAKEAWAY

Every technique in the table serves the same purpose: algebraically transforming a 0/0 indeterminate form into a function that can be directly evaluated. Think of it like clearing a fraction in algebra—you are not changing the function's behavior near the point, you are simply rewriting it in a form that reveals the limit value. Choosing the right technique is analogous to choosing the right tool in an engineering lab: a wrench and a socket both turn a bolt, but one fits the situation better.

SECTION 8

Connection to Advanced Topics

Removing discontinuities is not merely a stand-alone technique; it is foundational for several advanced topics in calculus. The very definition of the derivative relies on evaluating a limit that initially appears as a 0/0 indeterminate form. Understanding how to handle removable discontinuities prepares you for differentiation, L'Hôpital's Rule (covered in BC and college courses), and the analysis of functions defined by integrals.

Concept in This Lesson	Advanced Connection
Evaluating lim(x→a) [f(x) − f(a)]/(x − a) as a 0/0 form	This is the definition of f′(a). Every derivative computation is, in essence, removing a removable discontinuity of the difference quotient.
Factoring and cancelling common factors	L'Hôpital's Rule (AP Calculus BC) provides an alternative when algebraic simplification is intractable: differentiate numerator and denominator separately.
Redefining f(a) = L to achieve continuity	In real analysis, this idea generalizes to extensions of functions. A function continuous on (a, b) can sometimes be uniquely extended to [a, b] by assigning limit values at endpoints.
Piecewise functions with parameters	Differential equations often require matching conditions at boundaries—an extension of ensuring piecewise-defined solutions are continuous.

How removing discontinuities connects to advanced calculus

When you encounter the formal definition of the derivative in the next unit—f′(a) = lim(h→0) [f(a + h) − f(a)]/h—remember that plugging in h = 0 directly gives 0/0. You will need the same factoring, cancelling, and limit-evaluation skills you have practiced in this lesson. In that sense, mastering removable discontinuities is rehearsal for every differentiation problem to come.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A function f has a discontinuity at x = 3. Which of the following conditions is sufficient to guarantee that the discontinuity is removable?

A. f(3) is undefined
B. lim(x→3⁻) f(x) = lim(x→3⁺) f(x) = 5, and f(3) is defined
C. lim(x→3) f(x) exists and is finite
D. f(3) = lim(x→3⁺) f(x)
E. lim(x→3⁻) f(x) exists but lim(x→3⁺) f(x) does not exist

Show answer (C)

The correct answer is C. A discontinuity is removable if and only if the two-sided limit exists and is finite; this is the defining characteristic. Choice A is insufficient because f(3) being undefined does not tell us whether the limit exists—the function could have a vertical asymptote. Choice B describes a situation where f is already continuous (left limit = right limit = 5 and f(3) exists), so if f(3) = 5 there is no discontinuity, and if f(3) ≠ 5 the discontinuity is removable—but the condition as stated does not confirm a discontinuity exists. Choice D only matches f(3) to the right-hand limit, not the full two-sided limit, which would leave a possible jump. Choice E describes a situation where the two-sided limit fails to exist, making the discontinuity non-removable.

Question 2: BASIC CALCULATION

Let f(x) = (x² − 16)/(x − 4). What value should f(4) be defined as to make f continuous at x = 4?

A. 0
B. 4
C. 8
D. 16
E. The discontinuity cannot be removed

Show answer (C)

The correct answer is C. Factor the numerator: x² − 16 = (x − 4)(x + 4). For x ≠ 4, f(x) = (x − 4)(x + 4)/(x − 4) = x + 4. Therefore lim(x→4) f(x) = 4 + 4 = 8. Setting f(4) = 8 satisfies the continuity condition f(4) = lim(x→4) f(x). Choice A (0) would be f(4) if one mistakenly evaluated the original expression by plugging in 0/0 = 0. Choice B (4) confuses the input value with the output. Choice D (16) is the numerator value x² at x = 4, not the limit. Choice E is incorrect because the 0/0 form confirms the discontinuity is removable.

Question 3: INTERMEDIATE

Let g(x) = (√(x + 5) − 3)/(x − 4). What is lim(x→4) g(x)?

A. 1/6
B. 1/3
C. 3
D. 6
E. The limit does not exist

Show answer (A)

The correct answer is A. Direct substitution gives (√9 − 3)/(4 − 4) = 0/0, an indeterminate form, so we rationalize. Multiply numerator and denominator by the conjugate (√(x + 5) + 3): [(√(x + 5) − 3)(√(x + 5) + 3)] / [(x − 4)(√(x + 5) + 3)] = (x + 5 − 9) / [(x − 4)(√(x + 5) + 3)] = (x − 4) / [(x − 4)(√(x + 5) + 3)]. Cancel (x − 4) to get 1/(√(x + 5) + 3). As x → 4, this becomes 1/(√9 + 3) = 1/(3 + 3) = 1/6. Choice B (1/3) results from forgetting to include the +3 in the conjugate sum. Choice C (3) is √9, not the limit. Choice D (6) is the denominator of the correct answer, not the answer itself. Choice E is incorrect because the limit exists and is finite.

Question 4: APPLIED

4 pts

A piecewise function is defined as follows: f(x) = { (x³ − 8)/(x² − 4), x ≠ 2 and x ≠ −2 { k, x = 2 (a) Find lim(x→2) f(x). Show your algebraic work. (b) Determine the value of k that makes f continuous at x = 2. (c) Does f have any other discontinuities? Classify each as removable or non-removable. (d) If the discontinuity in part (c) is non-removable, explain why no single-point redefinition can fix it.

Show answer

Rubric (4 points): Part (a) — 1 point: Factor: x³ − 8 = (x − 2)(x² + 2x + 4) and x² − 4 = (x − 2)(x + 2). Cancel (x − 2): f(x) = (x² + 2x + 4)/(x + 2) for x ≠ 2, x ≠ −2. Evaluate: lim(x→2) f(x) = (4 + 4 + 4)/(2 + 2) = 12/4 = 3. Award 1 point for correct factoring and correct limit value of 3. Part (b) — 1 point: For continuity at x = 2, we need f(2) = lim(x→2) f(x), so k = 3. Award 1 point for stating k = 3 with justification referencing the definition of continuity. Part (c) — 1 point: The simplified form (x² + 2x + 4)/(x + 2) has a zero denominator at x = −2. The numerator at x = −2 is 4 − 4 + 4 = 4 ≠ 0, so the limit does not exist (it diverges). This is a non-removable (infinite) discontinuity / vertical asymptote at x = −2. Award 1 point for identifying x = −2 and correctly classifying it as non-removable/infinite. Part (d) — 1 point: At x = −2, the function approaches ±∞ from each side. Since the two-sided limit does not exist as a finite number, no single real number can be assigned to f(−2) to satisfy the definition of continuity. A removable discontinuity requires the existence of a finite limit. Award 1 point for a clear explanation connecting the non-existence of a finite limit to the impossibility of a single-point redefinition.

Question 5: CRITICAL THINKING

3 pts

Let h(x) = [sin(x² − π²)] / (x − π). Determine whether h has a removable discontinuity at x = π. If so, find the value that makes h continuous there. Justify your answer using limit techniques from this course.

Show answer

Rubric (3 points): Point 1 — Identifying the indeterminate form: At x = π, the numerator is sin(π² − π²) = sin(0) = 0 and the denominator is π − π = 0, giving a 0/0 indeterminate form. This indicates a potential removable discontinuity. Point 2 — Algebraic manipulation and limit evaluation: Factor x² − π² = (x − π)(x + π). Rewrite h(x) = sin[(x − π)(x + π)] / (x − π). Let u = (x − π)(x + π), so as x → π, u → 0. Use the identity sin(u)/u → 1 as u → 0. Write h(x) = {sin[(x − π)(x + π)] / [(x − π)(x + π)]} × (x + π). As x → π, the first factor approaches 1 and the second factor approaches 2π. Therefore lim(x→π) h(x) = 1 × 2π = 2π. Point 3 — Conclusion: Since the limit exists and equals 2π, the discontinuity at x = π is removable. Defining h(π) = 2π makes h continuous at x = π. Award this point for a clear statement that the discontinuity is removable and for the correct value 2π.

SUMMARY

Lesson Summary

Summary

A removable discontinuity occurs at x = a when the two-sided limit lim(x→a) f(x) = L exists and is finite, but the function either is undefined at a or takes a value different from L. To remove the discontinuity, redefine f(a) = L. This is the only type of discontinuity that can be repaired by a single-point redefinition; jump discontinuities and infinite discontinuities (vertical asymptotes) cannot be removed.

The algebraic techniques for evaluating the limit at a removable discontinuity include factoring and cancelling common factors, rationalization (multiplying by a conjugate), and applying trigonometric identities such as sin(u)/u → 1. On the AP exam, expect to encounter these in both multiple-choice and free-response contexts, including piecewise functions where you must determine a parameter value that ensures continuity at a specified point. The concept is also the algebraic foundation for the definition of the derivative, which is itself a limit of a 0/0 difference quotient.