AP Calculus BC • Limits and Continuity

Exploring Types of Discontinuities

Classifying the ways functions break reveals deep structure in limits and piecewise behavior.

AP Calculus BC illustration with summation, combined graphs, and an orbit
Inside this lesson 9 sections 5 questions 9 figures

SECTION 1

Historical Context & Motivation

The modern concept of a continuous function seems intuitive — draw a curve without lifting your pen — yet making this idea precise required centuries of effort. Early mathematicians like Euler freely manipulated formulas and assumed functions behaved "nicely" between computed values, but as analysts studied Fourier series and pathological constructions in the 1800s, they realized that a rigorous definition of continuity was essential. Understanding where and how functions fail to be continuous became just as important as understanding continuity itself, because those failures — discontinuities — encode information about limits, domain restrictions, and the algebraic structure of a function.

1821 • Cauchy's Cours d'Analyse

Augustin-Louis Cauchy published the first systematic treatment of limits and continuity, defining a function as continuous when an infinitely small increment in x produces an infinitely small increment in f(x).

1829 • Dirichlet's Function

Peter Gustav Lejeune Dirichlet introduced a function equal to 1 on rationals and 0 on irrationals — discontinuous everywhere — demonstrating that pathological discontinuities demanded rigorous classification.

1861 • Weierstrass's ε–δ Definition

Karl Weierstrass formalized continuity using the ε–δ framework, giving mathematicians a precise tool to prove or disprove continuity at every point and classify the type of failure when it occurs.

1875 • Darboux & the Intermediate Value Property

Jean-Gaston Darboux showed that derivatives always satisfy the intermediate value property, clarifying that derivatives can have discontinuities — but only of the second kind (essential), not removable or jump type.

The central question this lesson addresses is: when a function fails to be continuous at a point, what is the nature of that failure? On the AP Calculus BC exam, you must identify and classify discontinuities by analyzing one-sided limits, overall limits, and function values. The three classical types — removable, jump, and infinite (essential) — each arise from a different breakdown in the three-part definition of continuity, and recognizing them quickly is a foundational skill for everything from integration theory to series convergence.

SECTION 2

Core Principles & Definitions

A function f is continuous at x = c if and only if three conditions hold simultaneously: (1) f(c) is defined, (2) lim as x → c of f(x) exists, and (3) lim as x → c of f(x) = f(c). A discontinuity at x = c means at least one of these conditions fails. The specific condition that breaks determines the type of discontinuity.

1

Removable Discontinuity

The two-sided limit lim f(x) as x → c exists and is finite, but either f(c) is undefined or f(c) ≠ lim f(x). The "hole" can be "filled" by redefining f(c).
2

Jump Discontinuity

Both one-sided limits exist and are finite, but they are not equal: lim x→c⁻ f(x) ≠ lim x→c⁺ f(x). The function "jumps" from one value to another.
3

Infinite (Essential) Discontinuity

At least one of the one-sided limits is ±∞. The graph has a vertical asymptote at x = c. The two-sided limit does not exist in the real numbers.
4

Oscillating Discontinuity

The function oscillates between values infinitely often as x → c (e.g., sin(1/x) near x = 0). The limit does not exist due to unbounded oscillation rather than unbounded magnitude.

SECTION 3

Visual Explanation — Discontinuity Gallery

Three Classical Types of DiscontinuityRemovablexycf(c)limLimit exists;f(c) ≠ lim or undef.JumpxycL⁻L⁺Left limit ≠ Right limitInfinitexyc→ +∞+∞ ←Vertical asymptote
Figure 1. Left: a removable discontinuity — the open circle marks the limit, and the filled dot shows a misplaced f(c). Center: a jump discontinuity — the left-hand and right-hand limits (L⁻ and L⁺) are unequal. Right: an infinite discontinuity — both one-sided limits tend to +∞, producing a vertical asymptote (dashed amber line).

In Figure 1, each panel isolates the signature visual fingerprint of its discontinuity type. For the removable case, the curve approaches the same y-value from both sides (the open circle), but f(c) is either undefined or sits at a different height (the filled dot). Algebraically, this usually arises from a common factor that cancels, such as (x² − 4)/(x − 2) at x = 2. For the jump case, the curve has two finite landing heights that simply do not match; piecewise-defined functions are the classic source. Finally, the infinite case exhibits the telltale vertical asymptote: at least one branch of the graph shoots toward ±∞ as x approaches c, which is the hallmark of a zero in the denominator that does not cancel.

SECTION 4

Mathematical Framework

A precise classification of discontinuities follows directly from the ε–δ definition of continuity. We say f is continuous at c when for every ε > 0 there exists δ > 0 such that |x − c| < δ implies |f(x) − f(c)| < ε. When this fails, we examine the one-sided limits and the function value to determine the discontinuity type.

CONTINUITY CONDITION

fcontinuousatc⟺f(c)isdefined,lim(x→c)f(x)exists,andlim(x→c)f(x)=f(c)f continuous at c ⟺ f(c) is defined, lim(x→c) f(x) exists, and lim(x→c) f(x) = f(c)

All three conditions must hold. A discontinuity is classified by which condition(s) fail.

REMOVABLE DISCONTINUITY

lim(x→c−)f(x)=lim(x→c+)f(x)=Lbutf(c)≠Lorf(c)undefinedlim(x→c⁻) f(x) = lim(x→c⁺) f(x) = L but f(c) ≠ L or f(c) undefined

The two-sided limit L exists and is finite. Redefining f(c) = L "removes" the discontinuity.

JUMP DISCONTINUITY

lim(x→c−)f(x)=L−andlim(x→c+)f(x)=L+withL−≠L+(bothfinite)lim(x→c⁻) f(x) = L⁻ and lim(x→c⁺) f(x) = L⁺ with L⁻ ≠ L⁺ (both finite)

The one-sided limits both exist but differ. The size of the jump is |L⁺ − L⁻|. No redefinition of f(c) can make f continuous at c.

INFINITE DISCONTINUITY

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

At least one one-sided limit is unbounded. The graph features a vertical asymptote at x = c.

ℹ
AP EXAM NOTE

SECTION 5

Classification Flowchart & Decision Process

When faced with a function and asked to classify a discontinuity, a systematic decision tree is far more reliable than intuition alone. The flowchart below encodes the logical hierarchy: first check whether one-sided limits are finite, then whether they agree, and finally whether the function value matches the limit.

Discontinuity Classification Flowchartf is not continuous at x = cAre both one-sided limits finite?NoYesDoes limit oscillatewithout bound in value?No (±∞)YesINFINITEOSCILLATINGDo L⁻ and L⁺ equal each other?NoYesJUMPDoes f(c) = L?NoYesREMOVABLECONTINUOUSStart at the top. Follow the decision path based on limit behavior at x = c.
Figure 2. Decision flowchart: begin with whether one-sided limits are finite, then check equality of left- and right-hand limits, and finally verify the function value against the limit.

When working through the flowchart, the most common algebraic strategy is to factor and simplify the expression. If a common factor cancels and the remaining expression is defined at c, the original discontinuity is removable. If no factor cancels and the denominator is zero, the discontinuity is infinite. For piecewise functions, compute both one-sided limits directly by substituting into the appropriate piece; if they differ, the discontinuity is a jump.

SECTION 6

Worked Example

Let us classify every discontinuity of the function defined by:

GIVEN FUNCTION

g(x)=(x2−x−6)/(x2−9)g(x) = (x² − x − 6) / (x² − 9)

Determine the type of discontinuity at each point where g is not continuous.

Worked example

  1. 1

    Step 1 — Factor numerator and denominator

    Factor: x² − x − 6 = (x − 3)(x + 2) and x² − 9 = (x − 3)(x + 3). So g(x) = (x − 3)(x + 2) / [(x − 3)(x + 3)]. The domain excludes x = 3 and x = −3.
    g(x) = (x − 3)(x + 2) / [(x − 3)(x + 3)]
  2. 2

    Step 2 — Simplify and identify cancellation

    For x ≠ 3, the factor (x − 3) cancels, giving the simplified form g(x) = (x + 2)/(x + 3). The cancellation at x = 3 indicates a potential removable discontinuity; the remaining zero in the denominator at x = −3 indicates a potential infinite discontinuity.
    Simplified: g(x) = (x + 2)/(x + 3) for x ≠ 3, x ≠ −3
  3. 3

    Step 3 — Analyze x = 3 (removable?)

    Compute the two-sided limit: lim(x→3) (x + 2)/(x + 3) = 5/6. Since both one-sided limits equal 5/6 (finite and equal), but g(3) is undefined (zero in the original denominator), the three-part continuity test fails at condition (1). This is a removable discontinuity.
    x = 3: removable discontinuity; lim = 5/6
  4. 4

    Step 4 — Analyze x = −3 (infinite?)

    As x → −3⁻, the numerator approaches (−3 + 2) = −1, while the denominator (x + 3) → 0⁻, so g(x) → (−1)/(0⁻) = +∞. As x → −3⁺, the denominator → 0⁺, so g(x) → (−1)/(0⁺) = −∞. At least one (in fact both) one-sided limits are infinite, confirming an infinite discontinuity (vertical asymptote at x = −3).
    x = −3: infinite discontinuity (vertical asymptote)

SECTION 7

Comparing Discontinuity Types

FeatureRemovableJumpInfinite
Two-sided limit exists?Yes (finite)NoNo
One-sided limits finite?Yes, and equalYes, but unequalAt least one is ±∞
Can be "fixed"?Yes — redefine f(c)NoNo
Typical algebraic causeCommon factor cancelsPiecewise definitionZero in denominator (no cancel)
Graphical signatureHole (open circle)Gap between two branchesVertical asymptote
f integrable over [a, b] containing c?Yes (Riemann)Yes (finitely many jumps)Improper integral required

Comparison of the three main discontinuity types tested on AP Calculus BC

SECTION 8

Connection to Advanced Topics

Understanding discontinuities is not merely an exercise in classification — it has direct consequences throughout the AP Calculus BC curriculum and beyond. The Intermediate Value Theorem (IVT) requires continuity on a closed interval, so knowing where discontinuities lie tells you exactly where IVT conclusions are valid and where they may fail. Similarly, the Fundamental Theorem of Calculus requires the integrand to be continuous on [a, b]; when it is not, you must split the interval or invoke improper integral techniques.

Topic in This CourseHow Discontinuities Appear
Limits & Continuity (Unit 1)Classifying discontinuities from graphs and algebraic expressions; ε–δ arguments.
Differentiation (Units 2–3)A function must be continuous at a point to be differentiable there; discontinuity ⇒ non-differentiability.
Integration (Units 6–8)Removable and jump discontinuities are fine for Riemann integrability; infinite discontinuities require improper integrals.
Series (Unit 10)Power series define functions that are continuous inside their interval of convergence; behavior at endpoints may introduce discontinuities.
Parametric & Polar (Unit 9)Piecewise parametric paths can introduce jump discontinuities in dx/dt or dy/dt, affecting arc length and tangent computations.

Discontinuities thread through nearly every unit of AP Calculus BC.

In more advanced mathematics, the study of discontinuities connects to measure theory and Lebesgue integration, where the set of discontinuity points determines integrability (Lebesgue's criterion states that a bounded function is Riemann integrable if and only if its set of discontinuities has measure zero). While that level of formalism is beyond the BC exam, recognizing the foundational role of discontinuity classification motivates why we devote careful attention to it now.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Which of the following correctly describes a removable discontinuity at x = c? (A) The left-hand and right-hand limits both exist but are not equal. (B) The two-sided limit exists and is finite, but f(c) is either undefined or not equal to that limit. (C) At least one one-sided limit is ±∞. (D) The function oscillates infinitely near x = c. (E) The function is defined at x = c, so no discontinuity can exist.
  • A. The left-hand and right-hand limits both exist but are not equal.
  • B. The two-sided limit exists and is finite, but f(c) is either undefined or not equal to that limit.
  • C. At least one one-sided limit is ±∞.
  • D. The function oscillates infinitely near x = c.
  • E. The function is defined at x = c, so no discontinuity can exist.
Show answer (B)
Correct: (B). A removable discontinuity occurs precisely when the two-sided limit exists (both one-sided limits are equal and finite) but the function value at c either does not exist or disagrees with that limit. (A) describes a jump discontinuity, where one-sided limits are unequal. (C) describes an infinite discontinuity with a vertical asymptote. (D) describes an oscillating discontinuity such as sin(1/x) at x = 0. (E) is false because a function can be defined at c yet still be discontinuous there (e.g., f(c) exists but does not equal the limit).

Question 2: BASIC CALCULATION

Consider f(x) = (x² − 4)/(x − 2). What type of discontinuity, if any, does f have at x = 2? (A) f is continuous at x = 2. (B) Removable discontinuity (C) Jump discontinuity (D) Infinite discontinuity (E) Oscillating discontinuity
  • A. f is continuous at x = 2.
  • B. Removable discontinuity
  • C. Jump discontinuity
  • D. Infinite discontinuity
  • E. Oscillating discontinuity
Show answer (B)
Correct: (B). Factor: x² − 4 = (x − 2)(x + 2), so f(x) = (x − 2)(x + 2)/(x − 2) = x + 2 for x ≠ 2. The limit as x → 2 is 4, but f(2) is undefined (0/0). Since the two-sided limit exists and is finite but the function value is missing, this is a removable discontinuity. (A) is wrong because f(2) is undefined. (C) is wrong because both one-sided limits equal 4 (they agree). (D) is wrong because the limit is finite, not ±∞. (E) is wrong because there is no oscillatory behavior.

Question 3: INTERMEDIATE

Let h(x) = (x² + x − 6) / (x² − x − 2). Classify the discontinuities of h. (A) Removable at x = 2 and infinite at x = −1 (B) Infinite at x = 2 and removable at x = −1 (C) Removable at both x = 2 and x = −1 (D) Infinite at both x = 2 and x = −1 (E) Removable at x = −3 and infinite at x = 2
  • A. Removable at x = 2 and infinite at x = −1
  • B. Infinite at x = 2 and removable at x = −1
  • C. Removable at both x = 2 and x = −1
  • D. Infinite at both x = 2 and x = −1
  • E. Removable at x = −3 and infinite at x = 2
Show answer (A)
Correct: (A). Factor the numerator: x² + x − 6 = (x + 3)(x − 2). Factor the denominator: x² − x − 2 = (x − 2)(x + 1). The common factor (x − 2) cancels, leaving h(x) = (x + 3)/(x + 1) for x ≠ 2, x ≠ −1. At x = 2: lim = (2 + 3)/(2 + 1) = 5/3, finite, so the discontinuity is removable. At x = −1: the simplified form has (x + 1) in the denominator; as x → −1, the numerator → 2 while the denominator → 0, producing ±∞, so this is an infinite discontinuity. (B) reverses the types. (C) is wrong because x = −1 gives an unbounded limit. (D) is wrong because x = 2 has a finite limit. (E) misidentifies the points; x = −3 is not a point of discontinuity (it's a zero of the numerator, not the denominator).

Question 4: APPLIED

A piecewise function models the cost C(x) in dollars of x units of a resource: C(x) = 3x + 1 for x < 5, C(x) = 2x + k for x ≥ 5. (a) Find the value of k that makes C continuous at x = 5. (b) If instead k = 4, classify the discontinuity at x = 5 and compute the size of the jump.
Show answer
(a) For continuity at x = 5, we need lim(x→5⁻) C(x) = C(5). The left-hand limit is 3(5) + 1 = 16. We need C(5) = 2(5) + k = 10 + k = 16, so k = 6. [1 pt: left-hand limit = 16; 1 pt: setting 10 + k = 16; 1 pt: k = 6.] (b) With k = 4: lim(x→5⁻) C(x) = 16 and lim(x→5⁺) C(x) = C(5) = 2(5) + 4 = 14. Since both one-sided limits are finite but 16 ≠ 14, there is a jump discontinuity. The size of the jump is |16 − 14| = 2 dollars. [1 pt: identifying both one-sided limits; 1 pt: classifying as jump; 1 pt: computing jump size = 2.]

Question 5: CRITICAL THINKING

Let f(x) = (sin x) / x for x ≠ 0. (a) Prove that f has a removable discontinuity at x = 0 by evaluating the appropriate limit. (b) A student claims: "If a function has a removable discontinuity at x = c, then the function must be undefined at x = c." Provide a counterexample to disprove this claim.
Show answer
(a) We compute lim(x→0) (sin x)/x. This is a fundamental limit equal to 1 (provable via the squeeze theorem: cos x ≤ (sin x)/x ≤ 1 for x near 0). Since the two-sided limit exists and equals 1 (finite), but f(0) is undefined (0/0 form), conditions (1) and (3) of continuity fail. Thus f has a removable discontinuity at x = 0, removable by defining f(0) = 1. [1 pt: correct limit = 1; 1 pt: noting f(0) undefined; 1 pt: concluding removable discontinuity.] (b) Counterexample: Define g(x) = (sin x)/x for x ≠ 0 and g(0) = 5. Now g(0) is defined (it equals 5), but lim(x→0) g(x) = 1 ≠ 5 = g(0). The two-sided limit exists but does not equal the function value, so this is a removable discontinuity at a point where the function IS defined. The student's claim is false. [1 pt: valid counterexample with f(c) defined; 1 pt: verifying limit ≠ f(c); 1 pt: concluding the claim is disproved.]

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