Unit 1 of AP Calculus AB — Limits and Continuity — is where the course begins, and it deserves more attention than most students give it. Limits are not just an introductory formality. They are the rigorous foundation on which derivatives and integrals are built. Students who have a shaky understanding of limits find themselves struggling to interpret derivative and integral concepts correctly months later.
This guide covers the key limit concepts tested in AP Calculus AB, explains the reasoning behind each type, and provides original practice problems with complete worked solutions.
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What Is a Limit?
A limit asks: as x approaches a certain value, what value does f(x) approach? The notation is:
limx→a f(x) = L
This is read “the limit of f(x) as x approaches a is L.” Crucially, a limit describes the behavior of a function near a point — not necessarily at that point. A function does not need to be defined at x = a for the limit to exist.
Key Limit Concepts in AP Calculus AB
1. One-Sided Limits
The left-hand limit, limx→a⁻ f(x), is the value f(x) approaches as x approaches a from the left.
The right-hand limit, limx→a⁺ f(x), is the value f(x) approaches from the right.
The two-sided limit exists if and only if both one-sided limits exist and are equal.
2. Limit Laws
If limx→a f(x) = L and limx→a g(x) = M, then:
- lim [f(x) + g(x)] = L + M
- lim [f(x) · g(x)] = L · M
- lim [f(x)/g(x)] = L/M, provided M ≠ 0
- lim [c · f(x)] = c · L for any constant c
- lim [f(x)]ⁿ = Lⁿ
3. Direct Substitution
For polynomial and rational functions at points where the denominator ≠ 0, you can evaluate limits by direct substitution: plug a into f(x).
Example: limx→3 (x² − 2x + 1) = 9 − 6 + 1 = 4
4. Indeterminate Forms and Algebraic Techniques
When direct substitution gives 0/0, the limit may still exist but requires algebra to resolve. Common techniques:
- Factor and cancel. Example: limx→2 (x² − 4)/(x − 2) = lim (x+2)(x−2)/(x−2) = lim (x+2) = 4.
- Rationalize. Multiply numerator and denominator by the conjugate to eliminate a radical.
- Simplify compound fractions. Find a common denominator for the numerator and simplify.
5. Limits at Infinity
As x → ±∞, polynomial-divided-by-polynomial limits are evaluated by comparing the degrees of numerator and denominator:
- If degree of numerator < degree of denominator: limit = 0
- If degree of numerator = degree of denominator: limit = ratio of leading coefficients
- If degree of numerator > degree of denominator: limit = ±∞ (no finite limit)
6. Infinite Limits and Vertical Asymptotes
A function has a vertical asymptote at x = a if limx→a |f(x)| → ∞. This occurs when the denominator approaches 0 while the numerator does not.
7. The Squeeze Theorem
If g(x) ≤ f(x) ≤ h(x) near x = a, and lim g(x) = lim h(x) = L, then lim f(x) = L. The classic application: limx→0 x·sin(1/x) = 0 because |x·sin(1/x)| ≤ |x| and |x| → 0.
8. Continuity
A function f is continuous at x = a if all three conditions hold:
- f(a) is defined
- limx→a f(x) exists
- limx→a f(x) = f(a)
Discontinuities are classified as removable (hole), jump, or infinite (vertical asymptote).
AP Calculus AB Limits Practice Problems
Direct Substitution
1. Evaluate: limx→4 (3x² − 5x + 2)
Substitute x = 4: 3(16) − 5(4) + 2 = 48 − 20 + 2 = 30
2. Evaluate: limx→−1 (2x³ + x − 3)
2(−1)³ + (−1) − 3 = −2 − 1 − 3 = −6
Factoring / Indeterminate Forms
3. Evaluate: limx→3 (x² − 9)/(x − 3)
(x+3)(x−3)/(x−3) = x + 3. Limit = 3 + 3 = 6
4. Evaluate: limx→−2 (x² + 5x + 6)/(x + 2)
Factor numerator: (x+2)(x+3). Cancel: x+3. Limit = −2 + 3 = 1
5. Evaluate: limx→0 (√(x+4) − 2)/x
Multiply by conjugate: (√(x+4) − 2)(√(x+4) + 2) / x(√(x+4) + 2) = (x+4−4) / (x(√(x+4)+2)) = x / (x(√(x+4)+2)) = 1/(√(x+4)+2). At x = 0: 1/(2+2) = 1/4
Limits at Infinity
6. Evaluate: limx→∞ (4x² − 3)/(2x² + x)
Same degree; ratio of leading coefficients: 4/2 = 2
7. Evaluate: limx→∞ (5x + 1)/(x² − 4)
Numerator degree < denominator degree. Limit = 0
8. Evaluate: limx→−∞ (3x³ − x)/(x² + 2)
Numerator degree > denominator degree. Leading term 3x³/x² = 3x → −∞. Limit = −∞
One-Sided Limits and Continuity
9. Let f(x) = { x² + 1 if x < 2; 5 if x = 2; 3x − 1 if x > 2 }. Find limx→2⁻ f(x), limx→2⁺ f(x), and determine if f is continuous at x = 2.
Left-hand limit: (2²+1) = 5. Right-hand limit: 3(2)−1 = 5. f(2) = 5. All three equal 5 → f is continuous at x = 2.
10. Let g(x) = { 2x + 3 if x ≤ 1; x² + 2 if x > 1 }. Is g continuous at x = 1?
Left-hand: 2(1)+3 = 5. Right-hand: 1² + 2 = 3. Left ≠ Right → g has a jump discontinuity at x = 1.
Intermediate Value Theorem
11. f(x) = x³ − 2x − 5 is continuous on [1, 3]. Show that f has a zero in (1, 3).
f(1) = 1 − 2 − 5 = −6 < 0. f(3) = 27 − 6 − 5 = 16 > 0. By the IVT, since f is continuous and changes sign on [1,3], there exists c in (1,3) such that f(c) = 0.
What to Study Next After Limits
Unit 2 of AP Calculus AB — Differentiation: Definition and Fundamental Properties — builds directly on limits. The derivative is defined as:
f'(x) = limh→0 [f(x+h) − f(x)] / h
Students who are shaky on limit algebra will struggle to evaluate this definition. Mastering the limit laws and the algebraic techniques in this guide is the most important preparation you can do before moving forward.
AP Calculus AB Exam Overview
The AP Calculus AB exam is approximately 3 hours and 15 minutes long. Section I includes 45 multiple-choice questions (Part A: 30 questions, no calculator; Part B: 15 questions, graphing calculator required). Section II includes 6 free-response questions (Part A: 2 questions, graphing calculator required; Part B: 4 questions, no calculator). Scores are reported on a 1–5 scale. Many colleges grant credit for a score of 3 or higher, though policies vary. Always check with the specific institution.
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