College Board AP 2023 (style)-style practice: Section II (free response, no calculator). Let f(x) = x^2 + 1 for x ≤ 2 and f(x) = ax - 1 for x > 2. Find the value of a that makes f continuous at x = 2, and justify using the definition of continuity.

A 3-point continuity-condition question. (1 point) f(2) = 2^2 + 1 = 5 and _x → 2^- f(x) = 5. (1 point) For continuity the right limit must also equal 5: _x → 2^+ f(x) = a(2) - 1 = 2a - 1, set 2a - 1 = 5. (1 point) Solve: 2a = 6, so a = 3. With a = 3, all three conditions hold (f(2) defined, limit exists since both sides give 5, and they are equal), so f is continuous at x = 2.

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What exactly must be true for a function to be continuous at a single point?

Topic 1.11 Defining Continuity at a Point: state and apply the three-part definition of continuity at a point and test functions against it.

A focused answer to AP Calculus AB Topic 1.11, giving the three-part definition of continuity at a point and applying it to piecewise functions, including solving for a parameter that makes a function continuous.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
The three-part definition
Testing a point
Making a piecewise function continuous
How each condition can fail
Continuity and the rest of the unit

What this topic is asking

The College Board (Topic 1.11) wants you to state and use the three-part definition of continuity at a point. A function is continuous at $a$ exactly when its value, its limit, and the agreement of the two all line up. A favorite exam task is solving for a constant in a piecewise function so that it is continuous.

The three-part definition

Each condition rules out a failure type: (1) fails at a missing value, (2) fails at a jump or asymptote, and (3) fails at a hole where the value is plotted at the wrong height.

Testing a point

Making a piecewise function continuous

When a function is defined by different formulas on either side of a join $x = a$ , continuity forces the two pieces to meet. Set the left-hand limit, the right-hand limit, and the function value all equal at $a$ , then solve for any unknown constant. This is the most common exam application of the definition.

Solving for continuity at a join

Find the value of $k$ that makes $g(x)$ continuous at $x = 1$ , where $g(x) = kx + 2$ for $x \leq 1$ and $g(x) = x^2 + 3$ for $x > 1$ .

step 1 Left-hand value and limit

For $x \leq 1$ , $g(x) = kx + 2$ , so $g(1) = k(1) + 2 = k + 2$ and $\lim_{x \to 1^-} g(x) = k + 2$ .

step 2 Right-hand limit

For $x > 1$ , $g(x) = x^2 + 3$ , so $\lim_{x \to 1^+} g(x) = 1^2 + 3 = 4$ .

step 3 Set the pieces equal

Continuity needs $\lim_{x \to 1^-} g(x) = \lim_{x \to 1^+} g(x) = g(1)$ , so $k + 2 = 4$ .

step 4 Solve

$k = 2$ . With $k = 2$ , $g(1) = 4$ , the limit exists and equals $4$ , and all three conditions hold, so $g$ is continuous at $x = 1$ .

How each condition can fail

It is worth seeing exactly which kind of break each condition catches, because exam questions are often built around a single failing condition. Condition (1), " $f(a)$ is defined", fails at a point genuinely outside the domain - for example a hole where the formula gives $\frac{0}{0}$ - so there is no value plotted at all. Condition (2), "the limit exists", fails at a jump (unequal one-sided limits) or at a vertical asymptote (an infinite limit). Condition (3), "the limit equals the value", fails at a point where both a value and a limit exist but they sit at different heights, which is the classic removable discontinuity with a stray filled dot. Diagnosing which condition fails not only tells you the function is discontinuous but also classifies the discontinuity, linking this topic directly to Topic 1.10.

Continuity and the rest of the unit

Continuity at a point is the building block for continuity over an interval (Topic 1.12), for removing discontinuities (Topic 1.13), and for the Intermediate Value Theorem (Topic 1.16), all of which require the function to be continuous. It also connects forward to differentiability in Unit 2, where continuity is a necessary (but not sufficient) condition for a derivative to exist. The piecewise "solve for the constant" task is the single most common way the three-part definition is examined: you set the relevant one-sided limit equal to the function value at the join and solve. When two unknown constants appear, you usually need a second condition (often differentiability, met later in Unit 2) to pin both down, but for continuity alone one equation per join is enough.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)1 marksSection I (multiple choice, no calculator). A function

f

satisfies

\lim_{x \to 2} f(x) = 5

and

f(2) = 3

. Which condition for continuity at

x = 2

fails? (A)

f(2)

is defined (B)

\lim_{x \to 2} f(x)

exists (C)

\lim_{x \to 2} f(x) = f(2)

(D) None fails

Show worked answer →

The correct answer is (C).

Continuity at $x = 2$ requires three things: $f(2)$ defined, $\lim_{x \to 2} f(x)$ exists, and they are equal. Here $f(2) = 3$ is defined (so A holds) and the limit $5$ exists (so B holds), but $\lim_{x \to 2} f(x) = 5 \neq 3 = f(2)$ , so the third condition fails. The function is discontinuous at $x = 2$ (a removable discontinuity).

AP 2023 (style)3 marksSection II (free response, no calculator). Let

f(x) = x^2 + 1

for

x \leq 2

and

f(x) = ax - 1

for

x > 2

. Find the value of

a

that makes

f

continuous at

x = 2

, and justify using the definition of continuity.

Show worked answer →

A 3-point continuity-condition question.

(1 point) $f(2) = 2^2 + 1 = 5$ and $\lim_{x \to 2^-} f(x) = 5$ . (1 point) For continuity the right limit must also equal $5$ : $\lim_{x \to 2^+} f(x) = a(2) - 1 = 2a - 1$ , set $2a - 1 = 5$ . (1 point) Solve: $2a = 6$ , so $a = 3$ . With $a = 3$ , all three conditions hold ( $f(2)$ defined, limit exists since both sides give $5$ , and they are equal), so $f$ is continuous at $x = 2$ .

Related dot points

Sources & how we know this

AP Calculus AB and BC Course and Exam Description — College Board (2020)