Page 1 of 196
AP CALCULUS BC: "STUFF YOU MUST
KNOW COLD" EXAM QUESTION BANK
Table of Contents
1. Unit 1: Limits and Continuity
2. Unit 2: Differentiation — Definition and Fundamental Properties
3. Unit 3: Differentiation — Composite, Implicit, and Inverse Functions
4. Unit 4: Contextual Applications of Differentiation
5. Unit 5: Analytical Applications of Differentiation
6. Unit 6: Integration and Accumulation of Change
7. Unit 7: Differential Equations
8.Unit 8: Applications of Integration
9. Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions (BC Only)
10. Unit 10: Infinite Sequences and Series (BC Only)
Unit 1: Limits and Continuity
Question 1
Topic: Evaluating limits graphically
Learning objective: Estimate limits from graphs, including one-sided limits
Difficulty: Easy
The graph of a function f is shown below. At x = 2, there is a vertical asymptote. As x approaches
2 from the left, the graph decreases without bound. As x approaches 2 from the right, the graph
increases without bound.
,Page 2 of 196
What is the value of limₓ→₂ f(x)?
A. 0
B. −∞
C. ∞
D. Does not exist
Correct answer: D
Detailed rationale: The limit as x→2 requires the left-hand and right-hand limits to be equal.
Here, the left-hand limit is −∞ and the right-hand limit is +∞. Because these one-sided limits are
not equal (and neither is finite), the two-sided limit does not exist.
Why the other options are incorrect:
A: 0 would require the function to approach 0 from both sides, which contradicts the described
behavior.
B: −∞ describes only the left-hand limit, not the two-sided limit.
C: ∞ describes only the right-hand limit, not the two-sided limit.
Exam tip: For a two-sided limit to exist, the left-hand limit must equal the right-hand limit. If one
side goes to +∞ and the other to −∞, the limit does not exist.
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**Question 2**
Topic: Limit laws — algebraic evaluation
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Learning objective: Evaluate limits using algebraic manipulation
Difficulty: Easy
Evaluate limₓ→₃ (x² − 9)/(x − 3).
A. 0
B. 3
C. 6
D. Does not exist
Correct answer: C
Detailed rationale: Direct substitution gives 0/0, an indeterminate form. Factor the numerator:
x² − 9 = (x − 3)(x + 3). Cancel the common factor (x − 3) for x ≠ 3. The limit becomes limₓ→₃ (x +
3) = 6.
Why the other options are incorrect:
A: Substituting directly gives 0/0, not 0. The limit is not the value of the numerator alone.
B: 3 is the value of x, not the limit. Confusing input with output is a common error.
D: The limit exists because the removable discontinuity can be resolved by factoring.
Exam tip: When direct substitution yields 0/0, factor and cancel before re-evaluating. This is the
most common algebraic limit technique.
---
**Question 3**
, Page 4 of 196
Topic: One-sided limits and continuity
Learning objective: Determine continuity at a point using the three-part definition
Difficulty: Moderate
Let f(x) = { x² + 1, x < 2; 5, x = 2; 3x − 1, x > 2 }.
Which of the following is true about f at x = 2?
A. f is continuous at x = 2.
B. f has a removable discontinuity at x = 2.
C. f has a jump discontinuity at x = 2.
D. f has an infinite discontinuity at x = 2.
Correct answer: B
Detailed rationale: Check the three conditions for continuity. First, f(2) = 5, so the function is
defined. Second, limₓ→₂⁻ f(x) = 2² + 1 = 5 and limₓ→₂⁺ f(x) = 3(2) − 1 = 5. The two-sided limit
equals 5. Third, since limₓ→₂ f(x) = 5 = f(2), all conditions are satisfied. Therefore f is continuous.
Wait — the Correct answer is A, not B. The limit equals 5 and f(2) = 5. Let me re-check: lim from
left = 5, lim from right = 5, f(2) = 5. All equal. So f is continuous.
*Correct answer: A*
Detailed rationale (corrected): The left-hand limit is 5, the right-hand limit is 5, and f(2) = 5.
Since all three values are equal, f is continuous at x = 2.
Why the other options are incorrect:
AP CALCULUS BC: "STUFF YOU MUST
KNOW COLD" EXAM QUESTION BANK
Table of Contents
1. Unit 1: Limits and Continuity
2. Unit 2: Differentiation — Definition and Fundamental Properties
3. Unit 3: Differentiation — Composite, Implicit, and Inverse Functions
4. Unit 4: Contextual Applications of Differentiation
5. Unit 5: Analytical Applications of Differentiation
6. Unit 6: Integration and Accumulation of Change
7. Unit 7: Differential Equations
8.Unit 8: Applications of Integration
9. Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions (BC Only)
10. Unit 10: Infinite Sequences and Series (BC Only)
Unit 1: Limits and Continuity
Question 1
Topic: Evaluating limits graphically
Learning objective: Estimate limits from graphs, including one-sided limits
Difficulty: Easy
The graph of a function f is shown below. At x = 2, there is a vertical asymptote. As x approaches
2 from the left, the graph decreases without bound. As x approaches 2 from the right, the graph
increases without bound.
,Page 2 of 196
What is the value of limₓ→₂ f(x)?
A. 0
B. −∞
C. ∞
D. Does not exist
Correct answer: D
Detailed rationale: The limit as x→2 requires the left-hand and right-hand limits to be equal.
Here, the left-hand limit is −∞ and the right-hand limit is +∞. Because these one-sided limits are
not equal (and neither is finite), the two-sided limit does not exist.
Why the other options are incorrect:
A: 0 would require the function to approach 0 from both sides, which contradicts the described
behavior.
B: −∞ describes only the left-hand limit, not the two-sided limit.
C: ∞ describes only the right-hand limit, not the two-sided limit.
Exam tip: For a two-sided limit to exist, the left-hand limit must equal the right-hand limit. If one
side goes to +∞ and the other to −∞, the limit does not exist.
---
**Question 2**
Topic: Limit laws — algebraic evaluation
,Page 3 of 196
Learning objective: Evaluate limits using algebraic manipulation
Difficulty: Easy
Evaluate limₓ→₃ (x² − 9)/(x − 3).
A. 0
B. 3
C. 6
D. Does not exist
Correct answer: C
Detailed rationale: Direct substitution gives 0/0, an indeterminate form. Factor the numerator:
x² − 9 = (x − 3)(x + 3). Cancel the common factor (x − 3) for x ≠ 3. The limit becomes limₓ→₃ (x +
3) = 6.
Why the other options are incorrect:
A: Substituting directly gives 0/0, not 0. The limit is not the value of the numerator alone.
B: 3 is the value of x, not the limit. Confusing input with output is a common error.
D: The limit exists because the removable discontinuity can be resolved by factoring.
Exam tip: When direct substitution yields 0/0, factor and cancel before re-evaluating. This is the
most common algebraic limit technique.
---
**Question 3**
, Page 4 of 196
Topic: One-sided limits and continuity
Learning objective: Determine continuity at a point using the three-part definition
Difficulty: Moderate
Let f(x) = { x² + 1, x < 2; 5, x = 2; 3x − 1, x > 2 }.
Which of the following is true about f at x = 2?
A. f is continuous at x = 2.
B. f has a removable discontinuity at x = 2.
C. f has a jump discontinuity at x = 2.
D. f has an infinite discontinuity at x = 2.
Correct answer: B
Detailed rationale: Check the three conditions for continuity. First, f(2) = 5, so the function is
defined. Second, limₓ→₂⁻ f(x) = 2² + 1 = 5 and limₓ→₂⁺ f(x) = 3(2) − 1 = 5. The two-sided limit
equals 5. Third, since limₓ→₂ f(x) = 5 = f(2), all conditions are satisfied. Therefore f is continuous.
Wait — the Correct answer is A, not B. The limit equals 5 and f(2) = 5. Let me re-check: lim from
left = 5, lim from right = 5, f(2) = 5. All equal. So f is continuous.
*Correct answer: A*
Detailed rationale (corrected): The left-hand limit is 5, the right-hand limit is 5, and f(2) = 5.
Since all three values are equal, f is continuous at x = 2.
Why the other options are incorrect:
