AP Calculus BC, AP exams, 4–7%

Unit 1: Limits and Continuity

Unit 1 develops the limit and defines continuity in terms of it, working algebraically, graphically, and numerically. The squeeze theorem, asymptotic behavior, and the Intermediate Value Theorem complete the unit. In BC this material is weighted lightly on the exam, but it underlies the convergence reasoning in Unit 10.

Position: 1 of 10 units
Published weighting: 4–7% 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 BC. 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, one-sided limits, and when a limit fails to exist.
Determining limits algebraically: Substitution, factoring, rationalizing, and manipulating indeterminate forms.
Squeeze theorem and special limits: Bounding arguments and the standard trigonometric limits reused throughout the course.
Continuity: The three-part definition, classifying discontinuities, and continuity on an interval.
Asymptotes and limits at infinity: Relating vertical and horizontal asymptotes to infinite limits and end behavior.
Intermediate Value Theorem: Stating hypotheses and using the theorem to justify that a value is attained.

Study tips

How to practice Limits and Continuity.

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

Get limit notation exactly right now; it is required in Unit 10 convergence arguments where sloppiness costs points.
Practice limits at infinity for rational and exponential comparisons, since growth-rate reasoning recurs in series.
Rehearse stating theorem hypotheses before conclusions, which is how justification points are awarded.

Where this fits

Unit 1: Limits and Continuity is 1 of 10 units in the AP Calculus BC guide, and it carries a published weighting of 4–7% in the official framework. Use the full guide to balance it against everything else you have to review.

Back to the AP Calculus BC 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→0 of (1 - cos(x))/x?
1. A0
2. B1/2
3. C1
4. DThe limit does not exist
Show answer
Answer: A. 0
The form is 0/0, so L'Hospital's rule gives lim x→0 of sin(x)/1 = 0. The value 1/2 is the limit of (1 - cos(x))/x^2, which is the version students memorize more often, so it is the tempting distractor here.
Key insight: (1 - cos x)/x approaches 0, while (1 - cos x)/x^2 approaches 1/2; the exponent decides.
Question 2Foundational
The function f is defined by f(x) = 3x + 1 for x < 2 and f(x) = x^2 + c for x ≥ 2. What value of c makes f continuous at x = 2?
1. Ac = 3
2. Bc = -3
3. Cc = 7
4. Dc = 11
Show answer
Answer: A. c = 3
Continuity requires the one-sided limits to agree with the value: 3(2) + 1 = 7 must equal 2^2 + c = 4 + c, so c = 3. Choosing c = 7 sets the constant equal to the left-hand limit and forgets the 4 contributed by x^2.
Key insight: Match the one-sided limits at the break point, then solve for the unknown constant.
Question 3Intermediate
What is the value of lim x→0 of tan(x)/(5x)?
1. A1/5
2. B1
3. C5
4. D0
Show answer
Answer: A. 1/5
Write tan(x)/(5x) as (1/5) times (sin(x)/x) times (1/cos(x)). As x → 0 the middle factor approaches 1 and 1/cos(x) approaches 1, leaving 1/5. Answering 5 inverts the constant.
Key insight: Rewrite tangent as sine over cosine so the standard limit sin(x)/x can be applied.
Question 4Intermediate
Which theorem justifies the conclusion that lim x→0 of x^2 sin(1/x) = 0?
1. AThe squeeze theorem
2. BL'Hospital's rule
3. CThe Intermediate Value Theorem
4. DThe Mean Value Theorem
Show answer
Answer: A. The squeeze theorem
Because -1 ≤ sin(1/x) ≤ 1 for every nonzero x, the product is trapped between -x^2 and x^2, and both bounds approach 0 as x → 0. The squeeze theorem then forces the limit to be 0 even though sin(1/x) itself has no limit. L'Hospital's rule does not apply, because the expression is a product rather than an indeterminate quotient.
Key insight: A bounded factor times a factor going to zero has limit zero, by the squeeze theorem.
Question 5Advanced
What is the value of lim x→∞ of (√(4x^2 + 3x) - 2x)?
1. A0
2. B3/4
3. C3/2
4. DThe expression increases without bound
Show answer
Answer: B. 3/4
Multiply by the conjugate to get 3x/(√(4x^2 + 3x) + 2x), then divide numerator and denominator by x to get 3/(√(4 + 3/x) + 2), which approaches 3/(2 + 2) = 3/4. Answering 0 assumes the two growing terms cancel exactly, which the 3x term prevents.
Key insight: For an ∞ - ∞ form with radicals, multiply by the conjugate before taking the limit.

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Unit 1: Limits and Continuity