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AP Calculus BC 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

Let f, g, and h be functions satisfying g(x) ≤ f(x) ≤ h(x) for all x near c, except possibly at c. If lim (x→c) g(x) = 7 and lim (x→c) h(x) = 7, what is lim (x→c) f(x)?

All Questions (7)

Let f, g, and h be functions satisfying g(x) ≤ f(x) ≤ h(x) for all x near c, except possibly at c. If lim (x→c) g(x) = 7 and lim (x→c) h(x) = 7, what is lim (x→c) f(x)?

A) 0

B) 7

C) The limit cannot be determined from the information given.

D) The limit does not exist.

Correct Answer: B

According to the Squeeze Theorem, if a function f(x) is bounded between two other functions, g(x) and h(x), and both g(x) and h(x) approach the same limit L as x approaches c, then f(x) must also approach L. In this case, both bounding functions approach 7, so the limit of f(x) is 7.

Let f be a function such that for all x, 4x - x² ≤ f(x) ≤ x² - 4x + 8. What is the value of lim (x→2) f(x)?

A) 0

B) 2

C) 4

D) The limit cannot be determined.

Correct Answer: C

To solve this, we find the limits of the two bounding functions as x approaches 2. For the lower bound: lim (x→2) (4x - x²) = 4(2) - 2² = 8 - 4 = 4. For the upper bound: lim (x→2) (x² - 4x + 8) = 2² - 4(2) + 8 = 4 - 8 + 8 = 4. Since both the lower and upper bounds approach the same limit of 4, the Squeeze Theorem states that lim (x→2) f(x) must also be 4.

Which of the following is the value of lim (x→0) x²cos(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 the range of the cosine function is [-1, 1], so -1 ≤ cos(1/x) ≤ 1 for all x ≠ 0. Multiplying the inequality by x² (which is non-negative) gives -x² ≤ x²cos(1/x) ≤ x². As x approaches 0, both lim (x→0) (-x²) = 0 and lim (x→0) x² = 0. Therefore, by the Squeeze Theorem, the limit of the function squeezed between them must also be 0.

The Squeeze Theorem is used to determine the limit of a function f(x) as x approaches c by comparing it to two other functions, g(x) and h(x), such that g(x) ≤ f(x) ≤ h(x). What condition must be true for the theorem to apply?

A) lim (x→c) g(x) < lim (x→c) h(x)

B) lim (x→c) g(x) = lim (x→c) h(x)

C) f(c) must be equal to g(c) and h(c).

D) The functions g(x) and h(x) must be continuous at x = c.

Correct Answer: B

The fundamental condition for the Squeeze Theorem to determine a limit is that the two bounding functions must approach the exact same limit value. If lim (x→c) g(x) = L and lim (x→c) h(x) = L, then it can be concluded that lim (x→c) f(x) = L. If the limits are not equal, the theorem cannot be used to find the limit of f(x).

Suppose that for x > 5, the function f(x) satisfies the inequality (4x - 1)/x < f(x) < (4x² + 2x)/(x²). The limit of f(x) can be found as x approaches infinity by using the Squeeze Theorem. What is lim (x→∞) f(x)?

A) 0

B) 1

C) 4

D) The limit does not exist.

Correct Answer: C

We apply the Squeeze Theorem by finding the limits of the bounding functions as x approaches infinity. For the lower bound: lim (x→∞) (4x - 1)/x = lim (x→∞) (4 - 1/x) = 4. For the upper bound: lim (x→∞) (4x² + 2x)/x² = lim (x→∞) (4 + 2/x) = 4. Since both functions approach the limit 4, the limit of f(x) as x approaches infinity must also be 4.

If g(x) = x², h(x) = -x², and g(x) ≤ f(x) ≤ h(x) for all x, what can be concluded about lim (x→0) f(x)?

A) The limit is 0.

B) The limit is 1.

C) The limit does not exist.

D) The Squeeze Theorem does not apply because g(x) is not less than h(x).

Correct Answer: D

The premise for the Squeeze Theorem is that g(x) ≤ f(x) ≤ h(x). In this case, for any x ≠ 0, g(x) = x² is positive and h(x) = -x² is negative, so g(x) > h(x). The inequality g(x) ≤ h(x) is only true at x=0. Because the initial condition g(x) ≤ h(x) is not met for x near 0, the Squeeze Theorem cannot be applied as stated.

Let f(x) be a function such that lim (x→a) f(x) = 0. Which of the following limits can be determined to be 0 using the Squeeze Theorem, given the behavior of f(x)?

A) lim (x→a) [f(x) / sin(f(x))]

B) lim (x→a) [f(x) * sin(1/f(x))]

C) lim (x→a) [1 / f(x)]

D) lim (x→a) [sin(f(x))]

Correct Answer: B

This question requires recognizing the structure where the Squeeze Theorem is applicable. The expression in option B is of the form 'something going to zero' multiplied by 'something bounded'. We know -1 ≤ sin(θ) ≤ 1 for any θ, so -1 ≤ sin(1/f(x)) ≤ 1. We can create the inequality -|f(x)| ≤ f(x) * sin(1/f(x)) ≤ |f(x)|. Since we are given that lim (x→a) f(x) = 0, it follows that lim (x→a) |f(x)| = 0 and lim (x→a) -|f(x)| = 0. By the Squeeze Theorem, the limit of the expression in the middle must also be 0.