AP Calculus AB Practice Quiz: Determining Limits Using the Squeeze Theorem

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

Suppose that for all x in an interval containing c, g(x) ≤ f(x) ≤ h(x). If lim (x→c) g(x) = L and lim (x→c) h(x) = L, which of the following statements must be true according to the Squeeze Theorem?

A)lim (x→c) f(x) = LB)lim (x→c) f(x) does not exist.C)f(c) = LD)The limit of f(x) as x approaches c cannot be determined.

All Questions (7)

A) lim (x→c) f(x) = L

B) lim (x→c) f(x) does not exist.

C) f(c) = L

D) The limit of f(x) as x approaches c cannot be determined.

Correct Answer: A

The Squeeze Theorem states that if a function f(x) is 'squeezed' between two other functions, g(x) and h(x), and both g(x) and h(x) approach the same limit L at a certain point c, then f(x) must also approach that same limit L at point c.

Let f be a function such that 2x - 1 ≤ f(x) ≤ x² for all x. What is the value of lim (x→1) f(x)?

A) 0

B) 1

C) 2

D) The limit cannot be determined.

Correct Answer: B

We can use the Squeeze Theorem. First, find the limit of the lower bound: lim (x→1) (2x - 1) = 2(1) - 1 = 1. Next, find the limit of the upper bound: lim (x→1) x² = 1² = 1. Since the limits of both the lower and upper bounding functions are equal to 1, the limit of f(x) as x approaches 1 must also be 1.

What is lim (x→0) x²sin(1/x)?

A) -1

B) 0

C) 1

D) The limit does not exist.

Correct Answer: B

This limit can be found using the Squeeze Theorem. We know that the sine function is always bounded between -1 and 1, so -1 ≤ sin(1/x) ≤ 1 for all x ≠ 0. Multiplying the inequality by x² (which is non-negative) gives -x² ≤ x²sin(1/x) ≤ x². Now, we take the limits of the bounding functions as x approaches 0: lim (x→0) (-x²) = 0 and lim (x→0) (x²) = 0. Since both limits are 0, the limit of the function squeezed between them must also be 0.

A function f(x) satisfies the inequality 3x ≤ f(x) ≤ x³ + 2 for x ≥ 0. What conclusion can be drawn about lim (x→2) f(x) using the Squeeze Theorem?

A) The limit is 6.

B) The limit is 8.

C) The limit is 10.

D) No conclusion can be drawn because the limits of the bounding functions are not equal.

Correct Answer: D

To use the Squeeze Theorem, the limits of the upper and lower bounding functions must be equal. Let's evaluate them: lim (x→2) 3x = 3(2) = 6. The limit of the upper bound is lim (x→2) (x³ + 2) = 2³ + 2 = 8 + 2 = 10. Since 6 ≠ 10, the conditions for the Squeeze Theorem are not met, and we cannot determine the limit of f(x) using this method.

Let g be a function that satisfies the inequality (x² - 9)/(x - 3) ≤ g(x) ≤ x² - 2x + 3 for all x ≠ 3. What is lim (x→3) g(x)?

A) 3

B) 6

C) 9

D) The limit cannot be determined.

Correct Answer: B

We apply the Squeeze Theorem by finding the limits of the bounding functions as x approaches 3. For the lower bound, we use an equivalent expression by factoring: lim (x→3) (x² - 9)/(x - 3) = lim (x→3) (x - 3)(x + 3)/(x - 3) = lim (x→3) (x + 3) = 3 + 3 = 6. For the upper bound, we substitute directly: lim (x→3) (x² - 2x + 3) = 3² - 2(3) + 3 = 9 - 6 + 3 = 6. Since both limits are equal to 6, the limit of g(x) as x approaches 3 must be 6.

Which of the following is a required condition to determine the limit of a function f(x) as x approaches c using the Squeeze Theorem with bounding functions g(x) and h(x)?

A) f(x) must be a continuous function at x = c.

B) g(x) and h(x) must be equal at x = c.

C) The limits of g(x) and h(x) as x approaches c must exist and be equal.

D) f(x) must be differentiable at x = c.

Correct Answer: C

The core principle of the Squeeze Theorem is that if f(x) is trapped between g(x) and h(x), and both g(x) and h(x) approach the exact same value L as x approaches c, then f(x) has no choice but to also approach L. Therefore, the limits of the bounding functions must exist and be equal.

Let f be a function defined for all real numbers. If 5 - x⁴ ≤ f(x) ≤ 5 + cos(x) - 1/cos(x) for all x in the interval (-π/2, π/2), what is lim (x→0) f(x)?

A) 0

B) 4

C) 5

D) The limit does not exist.

Correct Answer: C

Using the Squeeze Theorem, we evaluate the limits of the two bounding functions as x approaches 0. For the lower bound: lim (x→0) (5 - x⁴) = 5 - 0⁴ = 5. For the upper bound: lim (x→0) (5 + cos(x) - 1/cos(x)) = 5 + cos(0) - 1/cos(0) = 5 + 1 - 1/1 = 5. Since both the lower and upper bounds approach the limit of 5, the limit of f(x) as x approaches 0 must also be 5.