Defining Continuity at a Point - AP Calculus BC

SECTION 1

Historical Context & Motivation

The concept of continuity lies at the very heart of calculus, yet its rigorous formulation was surprisingly late in arriving. For more than a century after Newton and Leibniz developed the core machinery of differentiation and integration, mathematicians relied on an informal, geometric intuition—a continuous function was simply one whose graph you could draw without lifting your pen. While this heuristic serves well for polynomials and trigonometric functions, the discovery of pathological functions in the early nineteenth century exposed its inadequacy and compelled analysts to craft the precise, limit-based definition that now anchors every introductory calculus course.

1687 • Newton's Principia

Isaac Newton published the Principia Mathematica, using "fluents" and "fluxions" to describe motion. Continuity was assumed intuitively—quantities were expected to vary smoothly through time.

1817 • Bolzano's Definition

Bernard Bolzano became one of the first to define continuity rigorously: a function is continuous at a point when the difference f(x + ω) − f(x) can be made smaller than any given quantity by taking ω sufficiently small. His work was largely overlooked during his lifetime.

1821 • Cauchy's Cours d'Analyse

Augustin-Louis Cauchy published his landmark textbook, systematizing the notion that continuity requires the limit of f(x) as x approaches a to equal f(a). This formulation became the backbone of real analysis.

1861 • Weierstrass's ε–δ Formalism

Karl Weierstrass introduced the modern epsilon-delta language during his Berlin lectures, removing all residual appeals to motion or geometric intuition and placing continuity on a purely arithmetic foundation.

1872 • Weierstrass's Everywhere-Continuous, Nowhere-Differentiable Function

Weierstrass shocked the mathematical community by constructing a function that is continuous at every point yet differentiable nowhere, demonstrating that continuity alone does not guarantee smoothness and highlighting the need for precise definitions.

The central question that drove these developments is deceptively simple: What does it mean, precisely, for a function to have no break at a specific point? Answering this question requires the language of limits, and the three-part definition of pointwise continuity that Cauchy and Weierstrass crystallized is exactly what the AP Calculus BC exam expects you to master.

SECTION 2

Core Principles & Definitions

A function f is said to be continuous at a point x = c if and only if three conditions are simultaneously satisfied. Each condition depends on the one before it, so failure at any stage means the function is discontinuous at c. This layered structure is not merely pedagogical—it gives you a systematic diagnostic tool for classifying every type of discontinuity you will encounter on the AP exam.

f(c) Is Defined

The function must actually assign a value at c. If c is not in the domain of f—because of division by zero, a square root of a negative number, or an explicit restriction—then the discussion of continuity at c cannot even begin.

lim x→c f(x) Exists

The two-sided limit of f(x) as x approaches c must exist and be finite. This requires that both the left-hand limit and the right-hand limit exist and are equal: lim x→c⁻ f(x) = lim x→c⁺ f(x).

lim x→c f(x) = f(c)

The limit value must agree with the function value. Even if f(c) exists and the limit exists, the function is discontinuous when these two values differ—a scenario that produces a removable discontinuity (hole) in the graph.

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KEY TAKEAWAY

Think of continuity at a point like arriving at a meeting: (1) the meeting room must exist, (2) all roads must lead to the same room, and (3) you must actually be in that room. If the room is missing, or the hallways lead to different rooms, or you ended up somewhere else, the meeting fails. Similarly, if f(c) is undefined, if the left and right limits disagree, or if the limit and the function value don't match, continuity fails.

It is worth noting that this definition is pointwise—it describes behavior at a single value c. A function is called continuous on an interval [a, b] when it is continuous at every interior point and satisfies the appropriate one-sided limit conditions at the endpoints. The distinction between pointwise and interval continuity becomes important when studying the Intermediate Value Theorem, the Extreme Value Theorem, and later, uniform continuity in analysis.

SECTION 3

Visualizing Continuity & Discontinuity

The diagram below contrasts a function that is continuous at x = c with three classic failure modes. Each panel illustrates which of the three conditions is violated, giving you a visual taxonomy of discontinuities.

Panel 1 (Continuous): The filled dot sits on the curve—limit equals function value. Panel 2 (Removable): The open circle marks the limit, but f(c) is the filled dot above it—condition 3 fails. Panel 3 (Jump): Left and right limits disagree—condition 2 fails. Panel 4 (Infinite): Both sides blow up toward infinity at the vertical asymptote—conditions 1 and 2 both fail.

Notice that the type of discontinuity directly corresponds to which condition is the first to fail when you check conditions 1, 2, and 3 in order. A removable discontinuity means conditions 1 and 2 pass but condition 3 fails (or condition 1 fails while the limit still exists). A jump discontinuity means condition 2 fails because the one-sided limits are unequal. An infinite discontinuity means the limit itself is unbounded, so condition 2 fails in a more dramatic way. Developing the habit of checking these conditions sequentially is one of the most reliable strategies for free-response questions on the AP exam.

SECTION 4

Mathematical Framework

We now translate the three verbal conditions into precise mathematical notation. The formal definition below is the version expected on the AP Calculus BC exam, and it connects naturally to the epsilon-delta framework you may encounter in a subsequent real-analysis course.

DEFINITION OF CONTINUITY AT A POINT

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

This single equation encapsulates all three conditions: f(c) must be defined (right side exists), lim(x→c) f(x) must exist (left side exists), and the two must be equal.

TWO-SIDED LIMIT EXISTENCE

lim(x→c) f(x) = L ⟺ lim(x→c⁻) f(x) = L and lim(x→c⁺) f(x) = L

The two-sided limit equals L only when both one-sided limits exist and agree. Here c⁻ denotes approach from the left (values less than c) and c⁺ denotes approach from the right (values greater than c).

EPSILON-DELTA FORMULATION

∀ ε > 0, ∃ δ > 0 such that 0 < |x − c| < δ ⟹ |f(x) − f(c)| < ε

This is the Weierstrass formalization. For every tolerance ε around f(c), there exists a neighborhood of radius δ around c that keeps all outputs within that tolerance. While the AP exam rarely asks you to construct ε–δ proofs, understanding this language deepens your grasp of continuity.

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AP Exam Tip

On the AP Calculus BC exam, continuity questions frequently appear in the context of piecewise-defined functions. You will be asked either to determine a value of a parameter that makes the function continuous or to identify at which point(s) the function is discontinuous. Always verify all three conditions, and show your work for each in free-response questions—partial credit depends on demonstrating that you checked each condition individually.

SECTION 5

Classifying Discontinuities

When a function fails to be continuous at a point, the nature of the failure determines the type of discontinuity. Understanding the classification is important both for communicating mathematical ideas precisely and for solving AP problems efficiently, since different types of discontinuities require different remediation strategies (e.g., a removable discontinuity can be "fixed" by redefining a single function value, whereas a jump or infinite discontinuity cannot).

Follow the flowchart from top to bottom. At each diamond-shaped decision node, check the relevant condition. The terminal rectangles (colored) identify the type of discontinuity—or confirm continuity when all conditions pass.

Discontinuity Type	Which Condition Fails	Graph Feature	Example
Removable	Condition 1 or Condition 3 (limit exists but ≠ f(c) or f(c) undefined)	A "hole" in the graph, possibly with a separate dot elsewhere	f(x) = (x² − 1)/(x − 1) at x = 1
Jump	Condition 2 (left-hand and right-hand limits both exist but are unequal)	A sudden vertical gap between two curve segments	f(x) = ⌊x⌋ (greatest integer function) at every integer
Infinite	Condition 2 (at least one one-sided limit is ±∞)	A vertical asymptote	f(x) = 1/x at x = 0
Oscillating	Condition 2 (limit does not exist due to infinite oscillation)	Infinitely rapid oscillation near c	f(x) = sin(1/x) at x = 0

Summary of Discontinuity Types

SECTION 6

Worked Example: Piecewise Continuity

Consider the piecewise function below. We will determine the value of the constant k that makes f continuous at x = 2, then verify all three conditions.

PIECEWISE FUNCTION

f(x) = { x² + k, if x < 2 { 3x - 1, if x \geq 2

The parameter k is a real constant. We need the function to transition seamlessly at x = 2.

Finding k for Continuity at x = 2

1
Step 1 — Check Condition 1: Is f(2) defined?

Since x = 2 falls in the domain x ≥ 2, we use the second piece: f(2) = 3(2) − 1 = 5. The function value exists, so Condition 1 is satisfied for any value of k.

f(2) = 5
2
Step 2 — Compute the right-hand limit

As x → 2⁺, we use the piece 3x − 1 (since x > 2 is covered by x ≥ 2). Thus lim(x→2⁺) f(x) = 3(2) − 1 = 5.

lim(x→2⁺) f(x) = 5
3
Step 3 — Compute the left-hand limit

As x → 2⁻, we use the piece x² + k (since x < 2). Thus lim(x→2⁻) f(x) = (2)² + k = 4 + k.

lim(x→2⁻) f(x) = 4 + k
4
Step 4 — Apply Condition 2: Set one-sided limits equal

For the two-sided limit to exist, we need the left-hand and right-hand limits to agree: 4 + k = 5, which gives k = 1.

k = 1
5
Step 5 — Verify Condition 3

With k = 1, the two-sided limit is lim(x→2) f(x) = 5, and f(2) = 5. Since the limit equals the function value, all three conditions are met, confirming that f is continuous at x = 2 when k = 1.

lim(x→2) f(x) = f(2) = 5 ✓

SECTION 7

Common Pitfalls & Exam Strategies

Students frequently lose points on continuity problems not because the underlying concept is difficult, but because they skip conditions or confuse related ideas. The table below catalogs the most common errors alongside corrective strategies that will save you points on both multiple-choice and free-response sections.

Common Pitfall	Why It's Wrong	Corrective Strategy
Assuming a function is continuous because the formula "looks nice"	Rational functions, piecewise functions, and functions with restricted domains can fail at specific points even when each piece is a polynomial	Always check domain restrictions and evaluate all three conditions at the point in question
Checking only one side of the limit	The two-sided limit requires agreement of both one-sided limits; a left-hand limit alone is insufficient	Compute lim x→c⁻ and lim x→c⁺ separately, then compare
Confusing "limit exists" with "function is defined"	A limit can exist at a point where the function is undefined (removable discontinuity), and the function can be defined where the limit does not exist	Treat conditions 1 and 2 as independent checks, not interchangeable
Plugging the boundary value into the wrong piece	In piecewise functions, the domain inequality determines which formula applies at the boundary point itself	Read the inequality carefully: ≤ or < makes a critical difference for f(c)
Forgetting to state conclusions explicitly on FRQs	The AP rubric awards points for a clear concluding statement connecting all three conditions	End with a sentence: "Since f(c) is defined, lim x→c f(x) exists, and lim x→c f(x) = f(c), f is continuous at x = c."

Common Pitfalls and Exam Strategies for Continuity Problems

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EXAM STRATEGY

Think of the three-condition check as a quality-control inspection on a manufacturing line: each station tests a different requirement, and the product (continuity) passes only if it clears every station. Skipping a station means a defective product might slip through. On the AP exam, each condition you explicitly verify is a potential rubric point earned—so always run the full inspection, even when you are confident of the answer.

SECTION 8

Connection to Advanced Theory

Pointwise continuity is the gateway to a family of deeper results that you will use throughout AP Calculus BC and beyond. Several of the most powerful theorems in calculus have continuity as a hypothesis, so being unable to verify continuity means being unable to apply these theorems. The table below sketches the broader landscape.

Concept	Relationship to Pointwise Continuity	Where It Appears
Continuity on an Interval	f is continuous on [a, b] if continuous at every interior point and satisfies one-sided conditions at the endpoints	IVT, EVT, FTC — all require interval continuity
Intermediate Value Theorem (IVT)	If f is continuous on [a, b] and N is between f(a) and f(b), then f(c) = N for some c in (a, b)	AP Calculus BC FRQ — justifying existence of roots
Extreme Value Theorem (EVT)	Continuous functions on a closed interval attain an absolute maximum and minimum	Optimization problems in Units 5 and 8
Differentiability Implies Continuity	If f is differentiable at c, then f is continuous at c. The converse is false (e.g., \|x\| at x = 0).	AP multiple-choice — logical implication questions
Fundamental Theorem of Calculus	Part 1 requires f to be continuous on [a, b] to guarantee that F(x) = ∫ₐˣ f(t) dt is differentiable with F′(x) = f(x)	Units 6 and 8 — accumulation functions

How Pointwise Continuity Connects to Major Theorems

In a real-analysis course, pointwise continuity generalizes further to uniform continuity, where a single δ works for all points in the domain simultaneously. The Heine–Cantor theorem guarantees that any function continuous on a closed bounded interval is automatically uniformly continuous—an elegant result that relies entirely on the pointwise definition you are learning now. For the AP Calculus BC exam, your immediate priority is mastering the three-condition test and recognizing its role as the gateway hypothesis for IVT, EVT, and the FTC.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Which of the following statements is the correct definition of continuity of a function f at a point x = c?

A. f(c) is defined and f is differentiable at c.
B. lim(x→c) f(x) exists.
C. f(c) is defined, lim(x→c) f(x) exists, and lim(x→c) f(x) = f(c).
D. The graph of f can be drawn without lifting the pen in some neighborhood of c.
E. f(c) is defined and lim(x→c) f(x) exists.

Show answer (C)

C is correct. The definition of continuity at x = c requires all three conditions: (1) f(c) is defined, (2) lim(x→c) f(x) exists, and (3) the limit equals the function value. A is wrong because differentiability is a stronger condition than continuity—differentiability implies continuity, but continuity does not require differentiability. B only covers condition 2 and ignores conditions 1 and 3; a function could have a limit at c yet be undefined there (removable discontinuity). D is an informal heuristic that is not mathematically rigorous and fails for many pathological functions. E omits the crucial third condition—f(c) could be defined and the limit could exist, yet f(c) ≠ lim(x→c) f(x).

Question 2: BASIC CALCULATION

Let g(x) = (x² − 9)/(x − 3) for x ≠ 3 and g(3) = 7. At x = 3, the function g has

A. a removable discontinuity because lim(x→3) g(x) = 6 ≠ g(3) = 7.
B. a jump discontinuity because the left-hand and right-hand limits differ.
C. an infinite discontinuity because x − 3 = 0 in the denominator.
D. no discontinuity; g is continuous at x = 3.
E. a removable discontinuity because g(3) is undefined.

Show answer (A)

A is correct. Factor the numerator: x² − 9 = (x − 3)(x + 3), so for x ≠ 3, g(x) = x + 3. Therefore lim(x→3) g(x) = 3 + 3 = 6. Since g(3) = 7 ≠ 6, condition 3 fails, producing a removable discontinuity. B is wrong because both one-sided limits equal 6, so they agree. C is wrong because the common factor cancels—the limit is finite, not infinite. D is wrong because the limit and function value disagree. E is wrong because g(3) is explicitly defined as 7; the discontinuity arises from a mismatch, not from an undefined value.

Question 3: INTERMEDIATE

The function h is defined by h(x) = { 2x + a, x < 1; x² + 3, x ≥ 1 }. For which value of a is h continuous at x = 1?

A. a = 1
B. a = 2
C. a = 3
D. a = 4
E. a = 0

Show answer (B)

B is correct. For continuity at x = 1, we need lim(x→1⁻) h(x) = h(1). Since x = 1 falls under x ≥ 1, h(1) = 1² + 3 = 4. The right-hand limit is also 4. For the left-hand limit: lim(x→1⁻) (2x + a) = 2(1) + a = 2 + a. Setting 2 + a = 4 gives a = 2. A (a = 1): left-hand limit = 3 ≠ 4. C (a = 3): left-hand limit = 5 ≠ 4. D (a = 4): left-hand limit = 6 ≠ 4. E (a = 0): left-hand limit = 2 ≠ 4.

Question 4: APPLIED

3 pts

A temperature sensor reports the temperature T(t) in °C at time t hours, where: T(t) = { 20 + 5sin(πt/6), 0 ≤ t < 6 { at + b, 6 ≤ t ≤ 12 } The engineers require T to be continuous at t = 6. Given that T(12) = 15, find the values of a and b. Justify your answer using the definition of continuity.

Show answer

Rubric (3 points): Point 1 — Evaluate T(6) and the left-hand limit: From the first piece, lim(t→6⁻) T(t) = 20 + 5sin(π·6/6) = 20 + 5sin(π) = 20 + 0 = 20. From the second piece, T(6) = 6a + b. For continuity, 6a + b = 20. Point 2 — Use the condition T(12) = 15 to form a second equation: T(12) = 12a + b = 15. Point 3 — Solve the system and state the continuity conclusion: Subtracting the first equation from the second: (12a + b) − (6a + b) = 15 − 20, so 6a = −5, giving a = −5/6. Substituting back: 6(−5/6) + b = 20 → −5 + b = 20 → b = 25. Since T(6) = 6(−5/6) + 25 = 20 = lim(t→6⁻) T(t), all three conditions for continuity are met at t = 6: T(6) is defined, the two-sided limit exists (both sides equal 20), and the limit equals T(6).

Question 5: CRITICAL THINKING

4 pts

Consider the function f(x) = { (x³ − 8)/(x² − 4), x < 2 { c, x = 2 { (√(x + 7) − 3)/(x − 2), x > 2 } (a) Determine whether lim(x→2) f(x) exists. If it exists, find its value. (b) Find the value of c that makes f continuous at x = 2, or explain why no such value exists. (c) Even with the optimal value of c from part (b), explain whether f is differentiable at x = 2.

Show answer

Rubric (4 points): Point 1 — Left-hand limit: Factor: x³ − 8 = (x − 2)(x² + 2x + 4) and x² − 4 = (x − 2)(x + 2). For x ≠ 2, f(x) = (x² + 2x + 4)/(x + 2). Thus lim(x→2⁻) f(x) = (4 + 4 + 4)/(4) = 12/4 = 3. Point 2 — Right-hand limit: Rationalize: multiply numerator and denominator by (√(x + 7) + 3). The numerator becomes (x + 7) − 9 = x − 2, so f(x) = (x − 2)/((x − 2)(√(x + 7) + 3)) = 1/(√(x + 7) + 3) for x ≠ 2. Thus lim(x→2⁺) f(x) = 1/(√9 + 3) = 1/6. Since 3 ≠ 1/6, the left- and right-hand limits disagree, so lim(x→2) f(x) does not exist. Point 3 — Continuity analysis: Because the two-sided limit does not exist (condition 2 fails), no value of c can make f continuous at x = 2. Continuity requires the two-sided limit to exist and equal f(2), but since the one-sided limits are unequal, this is impossible regardless of the choice of c. Point 4 — Differentiability: Differentiability at a point requires continuity at that point (since differentiability implies continuity). Since f cannot be made continuous at x = 2, f is necessarily not differentiable at x = 2 for any value of c. Even the closest choice (e.g., c = 3 or c = 1/6) only matches one side, leaving a jump discontinuity, which precludes differentiability.

SUMMARY

Lesson Summary

Summary

A function f is continuous at a point x = c if and only if three conditions hold: f(c) is defined, lim(x→c) f(x) exists (requiring both one-sided limits to agree), and the limit equals the function value. Failure at any stage produces a discontinuity that is classified as removable, jump, infinite, or oscillating depending on how the conditions break down.

This pointwise definition is the foundational hypothesis for the Intermediate Value Theorem, the Extreme Value Theorem, and the Fundamental Theorem of Calculus. On the AP exam, always verify all three conditions explicitly—especially in piecewise function problems—and connect your conclusion to the definition in a complete sentence to maximize rubric credit.