Exploring Types of Discontinuities - AP Calculus AB

SECTION 1

Historical Context & Motivation

The rigorous study of continuity and its failures emerged from centuries of mathematical debate about what it means for a function to behave "smoothly." Early practitioners of calculus — Newton and Leibniz among them — operated with an intuitive notion that functions were inherently continuous curves, but by the early nineteenth century, mathematicians began constructing examples that shattered this assumption. Functions with isolated holes, sudden jumps, and vertical asymptotes demanded a precise vocabulary and classification system, ultimately forging the modern epsilon-delta framework and the taxonomy of discontinuities that every calculus student studies today.

1821 • Cauchy's Cours d'Analyse

Augustin-Louis Cauchy published the first rigorous definition of continuity, stating that a function is continuous when an infinitely small increment in the variable always produces an infinitely small change in the function itself.

1829 • Dirichlet's Pathological Function

Peter Gustav Lejeune Dirichlet introduced a function equal to 1 on rationals and 0 on irrationals — discontinuous everywhere — forcing mathematicians to refine their definitions.

1861 • Weierstrass Formalizes Limits

Karl Weierstrass introduced the epsilon-delta (ε–δ) definition of a limit, providing the foundation for classifying exactly how and where functions fail to be continuous.

1870s • Classification of Discontinuities

Building on Weierstrass's framework, mathematicians established the standard taxonomy: removable, jump, and infinite (essential) discontinuities, a classification central to modern analysis and AP Calculus.

This historical arc raises the central question for our lesson: given a function that is not continuous at a point, how exactly does it fail? Identifying the type of discontinuity tells us whether the break is "fixable," represents a fundamental gap in the function's behavior, or signals an unbounded explosion — distinctions that matter profoundly when we later study derivatives, integrals, and the theorems (like the Intermediate Value Theorem) that require continuity as a hypothesis.

SECTION 2

Core Definitions & Principles

Before classifying discontinuities, recall the three-part definition of continuity at a point x = c: (1) f(c) is defined, (2) lim as x → c of f(x) exists, and (3) lim as x → c of f(x) equals f(c). A discontinuity occurs whenever at least one of these conditions fails. The specific condition (or conditions) that break determines the type of discontinuity.

Removable Discontinuity

The two-sided limit exists at x = c, but either f(c) is undefined or f(c) ≠ lim f(x). The "hole" can be filled by redefining f(c).

Jump Discontinuity

Both one-sided limits exist and are finite, but lim x→c⁻ f(x) ≠ lim x→c⁺ f(x). The function "jumps" from one value to another.

Infinite Discontinuity

At least one of the one-sided limits is ±∞. The graph has a vertical asymptote at x = c, and no finite limit exists.

Continuity Checklist

Always verify all three conditions: f(c) defined, lim exists, and lim equals f(c). Failure at any stage is a discontinuity.

SECTION 3

Visual Explanation — Graphing Discontinuities

The diagram below places all three discontinuity types side by side on a single coordinate system, making the visual distinctions immediately clear. Study how the behavior of the curve near the point of discontinuity differs in each case.

Left panel: Removable — the limit L exists (open circle) but f(c) is at a different point (solid dot). Center panel: Jump — left-hand limit L⁻ and right-hand limit L⁺ are finite but unequal. Right panel: Infinite — the graph rockets toward ±∞ near the vertical asymptote (V.A.) at x = c.

Notice that in the removable case the two pieces of the curve approach the same height — the limit exists — but the function value is either missing or placed elsewhere. In the jump case, each side of the curve levels off at a finite value, yet those values disagree. In the infinite case, the curve escapes the finite plane entirely, making any notion of a finite limit impossible.

SECTION 4

Mathematical Framework

We now express each discontinuity type in precise limit notation. Understanding these formal statements is essential for exam-level reasoning, especially when a problem asks you to justify why a function is or is not continuous at a point.

CONTINUITY AT x = c

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

This single statement implicitly requires: (1) f(c) exists, (2) the two-sided limit exists, and (3) they are equal.

REMOVABLE DISCONTINUITY

lim(x→c) f(x) = L ∈ ℝ, but f(c) ≠ L or f(c) is undefined

The limit L is a finite real number. The discontinuity is removed by defining (or redefining) f(c) = L.

JUMP DISCONTINUITY

lim(x→c⁻) f(x) = L⁻ ∈ ℝ, lim(x→c⁺) f(x) = L⁺ ∈ ℝ, L⁻ ≠ L⁺

Both one-sided limits are finite but disagree, so the two-sided limit does not exist. The size of the jump is |L⁺ − L⁻|.

INFINITE DISCONTINUITY

lim(x→c⁻) f(x) = ±∞ or lim(x→c⁺) f(x) = ±∞

At least one one-sided limit diverges. The line x = c is a vertical asymptote. The two-sided limit does not exist.

ℹ

AP Exam Tip

SECTION 5

Detailed Classification & Decision Flowchart

The flowchart below provides a systematic algorithm for classifying discontinuities. Starting from the top, you evaluate one-sided limits and follow the decision path. This procedural approach is especially helpful under exam time pressure, where a clear mental model prevents errors.

Decision flowchart: begin by checking whether one-sided limits exist finitely, then compare them. If they match, compare the common limit to f(c). Each terminal box is color-coded to match the discontinuity type used throughout this lesson.

Type	Two-Sided Limit	One-Sided Limits	f(c)	Graphical Feature
Removable	Exists (= L)	L⁻ = L⁺ = L	≠ L or undefined	Hole (open circle)
Jump	Does not exist	L⁻ ≠ L⁺, both finite	Equals one side, or other	Break / step
Infinite	Does not exist	At least one is ±∞	Usually undefined	Vertical asymptote

Summary comparison of the three discontinuity types

SECTION 6

Worked Example — Classifying Discontinuities

Consider the piecewise function below. We will classify every discontinuity using the decision framework from Section 5.

GIVEN FUNCTION

g(x) = { (x² − 4)/(x − 2), x < 2 ; 5, x = 2 ; 3x − 2, x > 2 }

Defined piecewise with a rational expression for x < 2, a point value at x = 2, and a linear rule for x > 2.

Worked example

1
Step 1 — Left-hand limit

For x < 2, g(x) = (x² − 4)/(x − 2). Factor the numerator: (x − 2)(x + 2)/(x − 2) = x + 2 for x ≠ 2. Therefore lim(x→2⁻) g(x) = 2 + 2 = 4.

Left-hand limit = 4
2
Step 2 — Right-hand limit

For x > 2, g(x) = 3x − 2. Substituting x = 2: lim(x→2⁺) g(x) = 3(2) − 2 = 4.

Right-hand limit = 4
3
Step 3 — Two-sided limit

Since both one-sided limits equal 4, the two-sided limit exists: lim(x→2) g(x) = 4.

lim(x→2) g(x) = 4
4
Step 4 — Compare to f(c)

The function value g(2) = 5, but the limit is 4. Since 5 ≠ 4, condition (3) of continuity fails.

Removable discontinuity at x = 2
5
Step 5 — How to remove it

Redefining g(2) = 4 would make the function continuous at x = 2. Graphically, this means moving the solid dot from (2, 5) to the hole at (2, 4).

SECTION 7

Comparing Discontinuity Types — Strengths & Limitations

Different discontinuity types have different consequences for the theorems and techniques of calculus. A removable discontinuity is a minor nuisance — the limit machinery still works — whereas jump and infinite discontinuities present more serious obstacles.

Feature	Removable	Jump	Infinite
Two-sided limit exists?	Yes	No	No
Can be "fixed"?	Yes — redefine f(c)	No	No
IVT applies?	After removal, yes	No — hypothesis fails	No — hypothesis fails
Derivative exists?	Possibly, after repair	No	No
Common algebraic cause	0/0 after substitution	Piecewise rules disagree	Nonzero / 0

Impact of each discontinuity type on calculus operations

SECTION 8

Connection to Advanced Theory

The discontinuity classification you learn in AP Calculus AB is the entry point to a richer taxonomy used in real analysis and topology. In those courses, removable and jump discontinuities are grouped together as discontinuities of the first kind (because both one-sided limits are finite), while infinite discontinuities and oscillating discontinuities (such as sin(1/x) near x = 0) are called discontinuities of the second kind. This distinction becomes important when studying Riemann integrability and measure theory.

Concept	AP Calculus AB	Real Analysis / Beyond
Discontinuity classes	Removable, jump, infinite	First kind (removable + jump) vs. second kind (infinite + oscillating)
Continuity definition	Three-condition check at a point	ε–δ definition; topological (preimage of open sets)
Integrability	Assumed for standard functions	A bounded function is Riemann integrable iff its discontinuities form a set of measure zero
Oscillating discontinuity	Not tested on AP exam	sin(1/x) at x = 0 — neither one-sided limit exists

AP Calculus AB vs. advanced analysis

For now, the three-type classification — removable, jump, and infinite — covers every discontinuity scenario you will encounter on the AP exam. Mastering this taxonomy prepares you to state precise hypotheses for the Intermediate Value Theorem, the Extreme Value Theorem, and later, the Fundamental Theorem of Calculus, all of which require continuity on a closed interval.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Which of the following is the defining characteristic that distinguishes a removable discontinuity from a jump discontinuity? (A) The function value f(c) is undefined at a removable discontinuity but defined at a jump discontinuity. (B) The two-sided limit exists at a removable discontinuity but does not exist at a jump discontinuity. (C) The graph has a vertical asymptote at a removable discontinuity but not at a jump discontinuity. (D) Both one-sided limits are infinite at a removable discontinuity but finite at a jump discontinuity. (E) The function is defined on only one side of a removable discontinuity but on both sides of a jump discontinuity.

A. The function value f(c) is undefined at a removable discontinuity but defined at a jump discontinuity.
B. The two-sided limit exists at a removable discontinuity but does not exist at a jump discontinuity.
C. The graph has a vertical asymptote at a removable discontinuity but not at a jump discontinuity.
D. Both one-sided limits are infinite at a removable discontinuity but finite at a jump discontinuity.
E. The function is defined on only one side of a removable discontinuity but on both sides of a jump discontinuity.

Show answer (B)

Correct answer: (B). At a removable discontinuity, both one-sided limits are equal, so the two-sided limit exists (even though it may not equal f(c)). At a jump discontinuity, the one-sided limits are both finite but unequal, so the two-sided limit does not exist. (A) is wrong because f(c) can be defined at a removable discontinuity — it simply doesn't match the limit. (C) is wrong because vertical asymptotes characterize infinite discontinuities, not removable ones. (D) is wrong because one-sided limits are finite at a removable discontinuity. (E) is wrong because there is no requirement about the domain being one-sided for either type.

Question 2: BASIC CALCULATION

Let f(x) = (x² − 9)/(x − 3). At x = 3, what type of discontinuity does f have? (A) f is continuous at x = 3. (B) Removable discontinuity with limit 6. (C) Removable discontinuity with limit 3. (D) Jump discontinuity. (E) Infinite discontinuity.

A. f is continuous at x = 3.
B. Removable discontinuity with limit 6.
C. Removable discontinuity with limit 3.
D. Jump discontinuity.
E. Infinite discontinuity.

Show answer (B)

Correct answer: (B). Factor: f(x) = (x − 3)(x + 3)/(x − 3) = x + 3 for x ≠ 3. The limit as x → 3 is 3 + 3 = 6, but f(3) is undefined (0/0), so the discontinuity is removable with limit 6. (A) is wrong because f(3) is undefined. (C) is wrong because the limit is 6, not 3. (D) is wrong because both one-sided limits agree at 6. (E) is wrong because there is no vertical asymptote — the factor cancels.

Question 3: INTERMEDIATE

Let h(x) = { 2x + 1, x < 1 ; 4, x = 1 ; x², x > 1 }. Classify every discontinuity of h and justify your classification. (A) Jump at x = 1, because lim x→1⁻ = 3 and lim x→1⁺ = 1. (B) Continuous at x = 1. (C) Removable at x = 1, because lim x→1⁻ = lim x→1⁺ = 3 but h(1) = 4. (D) Infinite at x = 1. (E) Jump at x = 1, because lim x→1⁻ = 3 and lim x→1⁺ = 2.

A. Jump at x = 1, because lim x→1⁻ = 3 and lim x→1⁺ = 1.
B. Continuous at x = 1.
C. Removable at x = 1, because lim x→1⁻ = lim x→1⁺ = 3 but h(1) = 4.
D. Infinite at x = 1.
E. Jump at x = 1, because lim x→1⁻ = 3 and lim x→1⁺ = 2.

Show answer (A)

Correct answer: (A). From the left: lim(x→1⁻) h(x) = 2(1) + 1 = 3. From the right: lim(x→1⁺) h(x) = (1)² = 1. Since 3 ≠ 1, the one-sided limits are finite but unequal, so h has a jump discontinuity at x = 1. (B) is wrong because the one-sided limits disagree, so h is not continuous. (C) is wrong because the one-sided limits are not equal; a removable discontinuity requires both one-sided limits to be equal. (D) is wrong because both one-sided limits are finite. (E) is wrong because lim(x→1⁺) h(x) = 1, not 2.

Question 4: APPLIED

A company's cost function is C(x) = { 5x, 0 ≤ x ≤ 100 ; 5x − 50, x > 100 }, where x is the number of units produced. The company claims the cost function is continuous because it is defined for all x ≥ 0. (a) Evaluate lim(x→100⁻) C(x) and lim(x→100⁺) C(x). (b) Classify the discontinuity at x = 100 and explain its real-world meaning. (c) Determine the value of the discount that would make C continuous at x = 100.

Show answer

(a) [1 pt] lim(x→100⁻) C(x) = 5(100) = 500. lim(x→100⁺) C(x) = 5(100) − 50 = 450. (b) [2 pts] Since 500 ≠ 450, the two-sided limit does not exist and the discontinuity is a jump discontinuity with jump size |500 − 450| = 50. In real-world terms, producing the 101st unit triggers a retroactive $50 discount, causing the total cost to drop suddenly — a pricing cliff that could create perverse incentives near the threshold. (c) [1 pt] For continuity, we need lim(x→100⁺) C(x) = C(100) = 500. Setting 5(100) − d = 500 gives d = 0. Alternatively, changing the rule for x > 100 to C(x) = 5x − 50 + 50 = 5x would make C continuous. The discount at x = 100 must be $0 (no discontinuous jump) or the piecewise rule must be adjusted so both sides meet at 500.

Question 5: CRITICAL THINKING

Let f(x) = (x² − 5x + 6)/(x² − 4). (a) Find all points of discontinuity of f. (b) Classify each discontinuity as removable, jump, or infinite. (c) For each removable discontinuity, state the value of the limit and give the "repaired" function value that would make f continuous there.

Show answer

(a) [1 pt] Factor: f(x) = (x − 2)(x − 3)/[(x − 2)(x + 2)]. Discontinuities occur where the denominator is zero: x = 2 and x = −2. (b) [1 pt] At x = 2: the factor (x − 2) cancels, leaving f(x) = (x − 3)/(x + 2) for x ≠ 2. Since lim(x→2) (x − 3)/(x + 2) = (−1)/4 = −1/4, a finite limit exists but f(2) is undefined. This is a removable discontinuity. At x = −2: After cancellation, f(x) = (x − 3)/(x + 2), and as x → −2, the numerator → −5 (nonzero) while the denominator → 0. Therefore at least one one-sided limit is ±∞, giving an infinite discontinuity (vertical asymptote). (c) [1 pt] The only removable discontinuity is at x = 2. Redefining f(2) = −1/4 makes f continuous there.