AP CALCULUS BC • LIMITS AND CONTINUITY

Confirming Continuity over an Interval

Rigorously verifying that a function has no breaks, jumps, or holes across an entire domain interval.

Historical Context & Motivation

The notion of a function being continuous — behaving without breaks or sudden jumps — has been central to analysis since the early development of calculus. Newton and Leibniz implicitly assumed continuity when computing derivatives and integrals, yet neither provided a formal definition. It was not until the nineteenth century that mathematicians encountered pathological functions (nowhere-differentiable curves, for instance) that demanded precise language for what continuity actually means, both at a point and across an interval.

1687
Newton's Principia
Newton uses 'fluent' quantities that vary continuously, relying on intuitive continuity without formal definitions.
1817
Bolzano's Intermediate Value Theorem
Bernard Bolzano provides one of the first rigorous proofs that a continuous function on a closed interval attains every value between its endpoints.
1821
Cauchy's Cours d'Analyse
Augustin-Louis Cauchy publishes the first systematic ε-δ framework for limits and continuity, formalizing the concept at a point.
1861
Weierstrass Refines ε-δ
Karl Weierstrass sharpens Cauchy's definitions and introduces uniform continuity, distinguishing pointwise from interval-wide behavior.

Why does confirming continuity over an interval matter for modern calculus students? Theorems you rely on daily — the Intermediate Value Theorem, the Extreme Value Theorem, and the Fundamental Theorem of Calculus — all require continuity on an interval as a hypothesis. If you cannot verify that hypothesis, you cannot invoke the theorem. This lesson equips you to rigorously confirm that a function is continuous on a specified interval.

Core Principles & Definitions

Before addressing intervals, recall the definition of continuity at a point: a function f is continuous at x = c if and only if (1) f(c) is defined, (2) lim as x → c of f(x) exists, and (3) that limit equals f(c). Continuity over an interval extends this pointwise check to every point in the interval, with special attention to the endpoints when the interval is closed.

1

Continuity at an Interior Point

For every point c in the open interval (a, b), the two-sided limit of f(x) as x → c must exist and equal f(c).
2

Right-Continuity at a Left Endpoint

At x = a (left endpoint of [a, b]), the right-hand limit lim x → a⁺ f(x) must equal f(a). The left-hand limit is irrelevant because x = a is the boundary.
3

Left-Continuity at a Right Endpoint

At x = b (right endpoint of [a, b]), the left-hand limit lim x → b⁻ f(x) must equal f(b). We only approach from within the interval.
4

Open vs. Closed Intervals

On an open interval (a, b), only two-sided interior checks are needed. On a closed interval [a, b], the one-sided endpoint checks are additionally required.
KEY TAKEAWAY
KEY TAKEAWAY

Visualizing Continuity & Discontinuity on an Interval

Left panel: a function continuous on the closed interval [a, b] — the curve can be drawn without lifting the pen, and both endpoints are filled (included). Right panel: a jump discontinuity at an interior point c breaks continuity on [a, b].

In the left panel, every point on the curve between a and b satisfies the three-part continuity test: the function value exists, the limit exists, and they agree. The filled circles at a and b represent the one-sided endpoint checks — right-continuous at a, left-continuous at b. In the right panel, the open circle at c indicates that the limit from the left and the limit from the right disagree (a jump discontinuity), violating condition (2) at that interior point. Even though the function is well-behaved everywhere else, this single failure prevents us from declaring continuity on the interval.

Mathematical Framework

We now formalize the ideas from Section 2. The three definitions below constitute the complete mathematical framework for confirming continuity on open, closed, and half-open intervals.

CONTINUITY AT AN INTERIOR POINT
f is continuous at c ∈ (a, b) ⟺ lim(x→c) f(x) = f(c)
This is the standard three-part test compressed into a single equality: it implicitly requires f(c) to be defined and the two-sided limit to exist.
CONTINUITY ON A CLOSED INTERVAL [a, b]
f continuous on [a, b] ⟺ (∀c ∈ (a,b): lim(x→c) f(x) = f(c)) ∧ lim(x→a⁺) f(x) = f(a) ∧ lim(x→b⁻) f(x) = f(b)
The conjunction (∧) requires all three parts: interior two-sided continuity, right-continuity at the left endpoint, and left-continuity at the right endpoint.
PROPERTIES PRESERVING CONTINUITY
If f and g are continuous at c, then f ± g, f · g, f/g (g(c) ≠ 0), and f ∘ g are continuous at c.
Sums, differences, products, quotients (with nonzero denominator), and compositions of continuous functions remain continuous. This lets you build arguments about complicated functions from simpler ones.
AP Exam Tip

Types of Discontinuities & How They Block Interval Continuity

When confirming continuity on an interval, you need to know exactly what can go wrong. There are three primary types of discontinuity, each violating a different part of the continuity definition. Recognizing them lets you target your analysis efficiently: check the points most likely to be problematic rather than blindly testing every value.

Three discontinuity types that prevent interval continuity. A removable discontinuity (left) can theoretically be patched by redefining f(c). A jump discontinuity (center) involves unequal one-sided limits. An infinite discontinuity (right) features a vertical asymptote.
Summary of the three discontinuity types
TypeWhat FailsCommon Cause
Removablef(c) is undefined or f(c) ≠ lim f(x)Common factor cancels (e.g., (x²−1)/(x−1) at x = 1)
Jumplim x→c⁻ f(x) ≠ lim x→c⁺ f(x)Piecewise definitions with mismatched branches
InfiniteLimit does not exist (unbounded)Vertical asymptote, e.g., 1/(x − 2) at x = 2

Worked Example: Piecewise Function on [−1, 4]

Consider the piecewise function f defined by f(x) = x² + 1 for x < 2, and f(x) = 3x − 1 for x ≥ 2. Determine whether f is continuous on the closed interval [−1, 4].

1
Step 1 — Identify the pieces and the potential trouble pointThe function has two branches: f(x) = x² + 1 on (−∞, 2) and f(x) = 3x − 1 on [2, ∞). Each branch is a polynomial, hence continuous on its own domain. The only place where continuity might fail is the junction point x = 2, which lies in the interior of [−1, 4].
Potential discontinuity at x = 2
2
Step 2 — Compute the left-hand limit at x = 2For x → 2⁻ we use the first branch: lim(x→2⁻) (x² + 1) = (2)² + 1 = 5.
lim(x→2⁻) f(x) = 5
3
Step 3 — Compute the right-hand limit at x = 2For x → 2⁺ we use the second branch: lim(x→2⁺) (3x − 1) = 3(2) − 1 = 5.
lim(x→2⁺) f(x) = 5
4
Step 4 — Evaluate f(2) and compareSince x = 2 falls in the branch x ≥ 2, f(2) = 3(2) − 1 = 5. Because the left-hand limit, right-hand limit, and function value all equal 5, f is continuous at x = 2.
f(2) = 5 = lim f(x) → continuous at x = 2
5
Step 5 — Check the endpoints of [−1, 4]At x = −1 (left endpoint), we need right-continuity. Since −1 < 2, f(−1) = (−1)² + 1 = 2 and lim(x→−1⁺) (x² + 1) = 2. These agree. At x = 4 (right endpoint), we need left-continuity. Since 4 ≥ 2, f(4) = 3(4) − 1 = 11 and lim(x→4⁻) (3x − 1) = 11. These agree.
f is continuous on [−1, 4]

Strategies & Common Pitfalls

Strategies and associated pitfalls when confirming continuity
StrategyWhen to UsePitfall to Avoid
Cite function type (polynomial, exponential, etc.)Non-piecewise functions whose domain covers the intervalForgetting domain restrictions — e.g., √x is not defined for x < 0
Check piecewise junctionsAny piecewise-defined functionOnly checking that limits agree but neglecting to verify f(c) matches
Use algebraic continuity theorems (sums, products, compositions)Complex expressions built from simple continuous functionsApplying quotient rule without verifying the denominator ≠ 0 on the interval
One-sided limits at endpointsClosed or half-open intervalsUsing a two-sided limit at an endpoint, which is outside the domain
KEY TAKEAWAY
EXAM STRATEGY

Connections to Major Theorems

Confirming continuity is rarely an end in itself; it is the gateway hypothesis for the most powerful results in calculus. The table below shows how interval continuity feeds directly into theorems you will use throughout the AP Calculus BC course. Failure to verify continuity before invoking any of these theorems is a logical error that the AP exam specifically tests.

Key calculus theorems requiring continuity on an interval
TheoremHypothesisConclusion
Intermediate Value Theorem (IVT)f continuous on [a, b]For any value L between f(a) and f(b), there exists c ∈ (a, b) with f(c) = L.
Extreme Value Theorem (EVT)f continuous on [a, b]f attains an absolute maximum and an absolute minimum on [a, b].
Fundamental Theorem of Calculus (FTC)f continuous on [a, b]∫ from a to b of f(x) dx = F(b) − F(a), where F′ = f.
Mean Value Theorem (MVT)f continuous on [a, b], differentiable on (a, b)∃ c ∈ (a, b) with f′(c) = (f(b) − f(a))/(b − a).

As you advance into topics like improper integrals, Taylor series convergence, and differential equations in Calculus BC, the requirement for continuity (or piecewise continuity) will recur. Mastering how to confirm continuity now builds the verification habit that will serve you across every major unit.

Practice Problems

1
Which of the following conditions is NOT required to confirm that f is continuous on the closed interval [a, b]?
2
Let g(x) = (x² − 9)/(x − 3) for x ≠ 3, and g(3) = 7. On which of the following intervals is g continuous?
3
Define h(x) = { 2x + k, for x ≤ 1; x² + 3, for x > 1 }. For what value of k is h continuous on [−2, 5]?
PROBLEM 4APPLIED
A temperature sensor records the temperature T(t) in °C of a chemical reaction, where t is time in minutes. The function T is defined by T(t) = { 20 + 5t, for 0 ≤ t < 4; t² + a, for 4 ≤ t ≤ 10 }. (a) Find the value of a that makes T continuous on [0, 10]. (b) Explain why continuity of T on [0, 10] guarantees that the reaction reaches exactly 50°C at some time in that interval.
PROBLEM 5CRITICAL THINKING
Let f be defined on [0, 6] by f(x) = { x sin(1/x), for 0 < x ≤ 6; 0, for x = 0 }. (a) Prove that f is continuous at x = 0. (b) Explain why f is continuous on (0, 6]. (c) Conclude that f is continuous on [0, 6]. (d) The Extreme Value Theorem applies to f on [0, 6]. Explain what it guarantees and discuss why verifying continuity was essential before making that claim.
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