AP Calculus AB, AP exams, 10–12%

Unit 1: Limits and Continuity

Unit 1 introduces the limit as the idea that makes calculus work, then defines continuity in terms of it. Algebraic, graphical, and numerical approaches to limits are all expected, along with the squeeze theorem, asymptotic behavior, and the Intermediate Value Theorem. Correct limit notation established here is required for the rest of the course.

Position: 1 of 8 units
Published weighting: 10–12% in the official framework
Topics covered: 6 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Calculus AB. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Defining and estimating limits: Reading limits from graphs and tables, and understanding one-sided limits and when a limit fails to exist.
Determining limits algebraically: Direct substitution, factoring, rationalizing, and using algebraic manipulation on indeterminate forms.
Squeeze theorem and special limits: Bounding a function between two others, and the standard trigonometric limits used repeatedly later.
Continuity: The three-part definition of continuity at a point, classifying discontinuities, and removable versus jump behavior.
Asymptotes and infinite limits: Connecting vertical and horizontal asymptotes to infinite limits and limits at infinity.
Intermediate Value Theorem: Stating the hypotheses and using the theorem to justify the existence of a solution on an interval.

Study tips

How to practice Limits and Continuity.

Section-specific guidance that follows from how this material is usually tested.

Write limit notation correctly on every problem; the free-response section deducts for notation even when the value is right.
Practice stating theorem hypotheses before conclusions, since the exam asks for justification rather than assertion.
Work limits from graphs and tables as well as algebraically, because all three representations appear.

Where this fits

Unit 1: Limits and Continuity is 1 of 8 units in the AP Calculus AB guide, and it carries a published weighting of 10–12% in the official framework. Use the full guide to balance it against everything else you have to review.

Back to the AP Calculus AB guide Official exam information

Sample Limits and Continuity questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

Question 1Foundational
What is the value of lim x→3 of (x^2 - 9)/(x - 3)?
1. A0
2. B3
3. C6
4. DThe limit does not exist
Show answer
Answer: C. 6
Direct substitution gives the indeterminate form 0/0, so factor the numerator as (x - 3)(x + 3) and cancel the common factor, leaving x + 3. Evaluating that at x = 3 gives 6. Reading 0/0 as the number 0 is the classic slip behind the first choice.
Key insight: The form 0/0 is not an answer; it is a signal to factor or rationalize first.
Question 2Foundational
The function f is defined by f(x) = (x^2 - 4)/(x - 2) for x ≠ 2 and by f(2) = k. What value of k makes f continuous at x = 2?
1. A0
2. B2
3. C4
4. DNo value of k makes f continuous at x = 2
Show answer
Answer: C. 4
Continuity at x = 2 requires f(2) to equal lim x→2 of f(x). Cancelling the common factor gives f(x) = x + 2 for x ≠ 2, so the limit is 4 and k must equal 4. The discontinuity is removable, which is exactly why some value of k works.
Key insight: A removable discontinuity is repaired by defining the value to be the limit.
Question 3Intermediate
What is the value of lim x→∞ of (3x^2 - 5x + 1)/(2x^2 + 7)?
1. A0
2. B3/2
3. C3
4. DThe limit does not exist
Show answer
Answer: B. 3/2
Numerator and denominator have the same degree, so the limit at infinity is the ratio of the leading coefficients, 3/2. Dividing every term by x^2 makes this explicit because the -5x, 1, and 7 terms all vanish. Choosing 3 keeps the leading coefficient but forgets the denominator's 2.
Key insight: Equal degrees give a horizontal asymptote at the ratio of leading coefficients.
Question 4Intermediate
The function f is continuous on the closed interval [1, 5] with f(1) = -3 and f(5) = 7. What does the Intermediate Value Theorem guarantee?
1. AThere is at least one c in (1, 5) with f(c) = 0
2. BThere is exactly one c in (1, 5) with f(c) = 0
3. Cf is increasing on the interval [1, 5]
4. DThere is a c in (1, 5) with f'(c) = 2.5
Show answer
Answer: A. There is at least one c in (1, 5) with f(c) = 0
Because f is continuous and 0 lies between f(1) = -3 and f(5) = 7, the Intermediate Value Theorem guarantees at least one c in the open interval where f(c) = 0. It never promises uniqueness or monotonicity, and the statement about f'(c) is the Mean Value Theorem, which also requires differentiability.
Key insight: The Intermediate Value Theorem gives existence, not uniqueness or a derivative value.
Question 5Advanced
What is the value of lim x→0 of sin(5x)/(3x)?
1. A0
2. B3/5
3. C1
4. D5/3
Show answer
Answer: D. 5/3
Rewrite the quotient as (5/3) times sin(5x)/(5x). As x → 0 the factor sin(5x)/(5x) approaches 1 by the standard trigonometric limit, so the value is 5/3. Answering 3/5 inverts the constants, and answering 1 applies the standard limit without tracking them at all.
Key insight: Match the argument of the sine to the denominator before using lim sin(u)/u = 1.

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Unit 1: Limits and Continuity