AP Calculus AB Quiz: Determining Limits Using The Squeeze Theorem
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Determining Limits Using The Squeeze TheoremQuestion 1 of 16

Which of the following is a correct application of the Squeeze Theorem to find lim⁡x→0x2sin⁡(1x)\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)?

Since −1≤sin⁡(1x)≤1-1 \leq \sin\left(\frac{1}{x}\right) \leq 1, we have −x2≤x2sin⁡(1x)≤x2-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2
Since sin⁡(1x)\sin\left(\frac{1}{x}\right) oscillates, the limit does not exist by the Squeeze Theorem
Since 0≤sin⁡(1x)≤10 \leq \sin\left(\frac{1}{x}\right) \leq 1, we have 0≤x2sin⁡(1x)≤x20 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2
The Squeeze Theorem cannot be applied because sin⁡(1x)\sin\left(\frac{1}{x}\right) is undefined at x=0x = 0
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AP Calculus AB Quiz

AP Calculus AB Quiz: Determining Limits Using The Squeeze Theorem

Practice Determining Limits Using The Squeeze Theorem in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Determining Limits Using The Squeeze Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Which of the following is a correct application of the Squeeze Theorem to find lim⁡x→0x2sin⁡(1x)\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)?

  1. Since −1≤sin⁡(1x)≤1-1 \leq \sin\left(\frac{1}{x}\right) \leq 1, we have −x2≤x2sin⁡(1x)≤x2-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2 (correct answer)
  2. Since sin⁡(1x)\sin\left(\frac{1}{x}\right) oscillates, the limit does not exist by the Squeeze Theorem
  3. Since 0≤sin⁡(1x)≤10 \leq \sin\left(\frac{1}{x}\right) \leq 1, we have 0≤x2sin⁡(1x)≤x20 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2
  4. The Squeeze Theorem cannot be applied because sin⁡(1x)\sin\left(\frac{1}{x}\right) is undefined at x=0x = 0
Explanation: Since −1≤sin⁡(1x)≤1-1 \leq \sin\left(\frac{1}{x}\right) \leq 1 for all x≠0x \neq 0, multiplying by x2≥0x^2 \geq 0 gives −x2≤x2sin⁡(1x)≤x2-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2. Since both lim⁡x→0(−x2)=0\lim_{x \to 0} (-x^2) = 0 and lim⁡x→0x2=0\lim_{x \to 0} x^2 = 0, the Squeeze Theorem gives lim⁡x→0x2sin⁡(1x)=0\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0.

Question 2

If ∣f(x)−3∣≤x2|f(x) - 3| \leq x^2 for all xx near 0, what is lim⁡x→0f(x)\lim_{x \to 0} f(x)?

  1. 00
  2. 33 (correct answer)
  3. −3-3
  4. The limit cannot be determined from this information
Explanation: The inequality ∣f(x)−3∣≤x2|f(x) - 3| \leq x^2 is equivalent to −x2≤f(x)−3≤x2-x^2 \leq f(x) - 3 \leq x^2, which gives 3−x2≤f(x)≤3+x23 - x^2 \leq f(x) \leq 3 + x^2. Since lim⁡x→0(3−x2)=3\lim_{x \to 0} (3 - x^2) = 3 and lim⁡x→0(3+x2)=3\lim_{x \to 0} (3 + x^2) = 3, the Squeeze Theorem gives lim⁡x→0f(x)=3\lim_{x \to 0} f(x) = 3.

Question 3

If 1−2∣x∣≤g(x)≤1+2∣x∣1 - 2|x| \leq g(x) \leq 1 + 2|x| for all xx near 0, what is lim⁡x→0g(x)\lim_{x \to 0} g(x)?

  1. 00
  2. 11 (correct answer)
  3. 22
  4. −1-1
Explanation: Since lim⁡x→0(1−2∣x∣)=1−2(0)=1\lim_{x \to 0} (1 - 2|x|) = 1 - 2(0) = 1 and lim⁡x→0(1+2∣x∣)=1+2(0)=1\lim_{x \to 0} (1 + 2|x|) = 1 + 2(0) = 1, and 1−2∣x∣≤g(x)≤1+2∣x∣1 - 2|x| \leq g(x) \leq 1 + 2|x|, the Squeeze Theorem gives lim⁡x→0g(x)=1\lim_{x \to 0} g(x) = 1. Both bounding functions approach the same limit of 1.

Question 4

If −x2≤f(x)≤x2-x^2 \leq f(x) \leq x^2 for all xx near 0, what is lim⁡x→0f(x)\lim_{x \to 0} f(x)?

  1. 00 (correct answer)
  2. 11
  3. −1-1
  4. The limit does not exist
Explanation: By the Squeeze Theorem, since lim⁡x→0(−x2)=0\lim_{x \to 0} (-x^2) = 0 and lim⁡x→0x2=0\lim_{x \to 0} x^2 = 0, and −x2≤f(x)≤x2-x^2 \leq f(x) \leq x^2, we have lim⁡x→0f(x)=0\lim_{x \to 0} f(x) = 0. The function f(x)f(x) is squeezed between two functions that both approach 0.

Question 5

What is lim⁡x→0x2+x4cos⁡(1x)\lim_{x \to 0} \sqrt{x^2 + x^4} \cos\left(\frac{1}{x}\right) using the Squeeze Theorem?

  1. 11
  2. 00 (correct answer)
  3. −1-1
  4. The limit does not exist due to oscillation
Explanation: Since −1≤cos⁡(1x)≤1-1 \leq \cos\left(\frac{1}{x}\right) \leq 1 and x2+x4≥0\sqrt{x^2 + x^4} \geq 0, we have −x2+x4≤x2+x4cos⁡(1x)≤x2+x4-\sqrt{x^2 + x^4} \leq \sqrt{x^2 + x^4} \cos\left(\frac{1}{x}\right) \leq \sqrt{x^2 + x^4}. Since x2+x4=∣x∣1+x2\sqrt{x^2 + x^4} = |x|\sqrt{1 + x^2} and lim⁡x→0∣x∣1+x2=0⋅1=0\lim_{x \to 0} |x|\sqrt{1 + x^2} = 0 \cdot 1 = 0, both bounds approach 0, so by the Squeeze Theorem, the limit is 0.

Question 6

If cos⁡x≤f(x)≤1\cos x \leq f(x) \leq 1 for all xx near π2\frac{\pi}{2}, what can be concluded about lim⁡x→π2f(x)\lim_{x \to \frac{\pi}{2}} f(x)?

  1. The limit equals 00 by the Squeeze Theorem since both bounds approach 00
  2. The limit equals 11 by the Squeeze Theorem since both bounds approach 11
  3. The limit cannot be determined by the Squeeze Theorem since the bounds approach different values (correct answer)
  4. The limit equals 12\frac{1}{2} by the Squeeze Theorem as the average of the bounds
Explanation: At x=π2x = \frac{\pi}{2}, cos⁡(π2)=0\cos\left(\frac{\pi}{2}\right) = 0 and the upper bound is 11. Since lim⁡x→π2cos⁡x=0\lim_{x \to \frac{\pi}{2}} \cos x = 0 and lim⁡x→π21=1\lim_{x \to \frac{\pi}{2}} 1 = 1, the bounding functions approach different limits (00 and 11). The Squeeze Theorem cannot be applied because both bounds must approach the same value.

Question 7

Using the Squeeze Theorem, what is lim⁡θ→0θ2cos⁡(1θ)\lim_{\theta \to 0} \theta^2 \cos\left(\frac{1}{\theta}\right)?

  1. 11
  2. 00 (correct answer)
  3. −1-1
  4. The limit oscillates and does not exist
Explanation: Since −1≤cos⁡(1θ)≤1-1 \leq \cos\left(\frac{1}{\theta}\right) \leq 1 for all θ≠0\theta \neq 0, multiplying by θ2≥0\theta^2 \geq 0 gives −θ2≤θ2cos⁡(1θ)≤θ2-\theta^2 \leq \theta^2 \cos\left(\frac{1}{\theta}\right) \leq \theta^2. Since lim⁡θ→0(−θ2)=0\lim_{\theta \to 0} (-\theta^2) = 0 and lim⁡θ→0θ2=0\lim_{\theta \to 0} \theta^2 = 0, the Squeeze Theorem gives lim⁡θ→0θ2cos⁡(1θ)=0\lim_{\theta \to 0} \theta^2 \cos\left(\frac{1}{\theta}\right) = 0.

Question 8

If ∣g(x)∣≤3x2|g(x)| \leq 3x^2 for all xx near 0, what is lim⁡x→0g(x)\lim_{x \to 0} g(x)?

  1. 33
  2. 00 (correct answer)
  3. −3-3
  4. The limit could be any value between −3-3 and 33
Explanation: The inequality ∣g(x)∣≤3x2|g(x)| \leq 3x^2 is equivalent to −3x2≤g(x)≤3x2-3x^2 \leq g(x) \leq 3x^2. Since lim⁡x→0(−3x2)=0\lim_{x \to 0} (-3x^2) = 0 and lim⁡x→03x2=0\lim_{x \to 0} 3x^2 = 0, the Squeeze Theorem gives lim⁡x→0g(x)=0\lim_{x \to 0} g(x) = 0. The absolute value inequality forces g(x)g(x) to be squeezed to 0.

Question 9

Which condition is essential for applying the Squeeze Theorem to find lim⁡x→af(x)\lim_{x \to a} f(x)?

  1. The functions g(x)g(x) and h(x)h(x) must be continuous at x=ax = a
  2. The function f(x)f(x) must be defined at x=ax = a
  3. The limits lim⁡x→ag(x)\lim_{x \to a} g(x) and lim⁡x→ah(x)\lim_{x \to a} h(x) must exist and be equal (correct answer)
  4. The inequality g(x)≤f(x)≤h(x)g(x) \leq f(x) \leq h(x) must hold for all real numbers
Explanation: The Squeeze Theorem requires that g(x)≤f(x)≤h(x)g(x) \leq f(x) \leq h(x) in some neighborhood of aa (not necessarily at aa itself), and that lim⁡x→ag(x)=lim⁡x→ah(x)=L\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L for some value LL. Then lim⁡x→af(x)=L\lim_{x \to a} f(x) = L. The bounding functions need not be continuous, ff need not be defined at aa, and the inequality need only hold near aa.

Question 10

Which of the following inequalities would allow us to conclude that lim⁡x→0f(x)=0\lim_{x \to 0} f(x) = 0 using the Squeeze Theorem?

  1. −x4≤f(x)≤x2-x^4 \leq f(x) \leq x^2 for all xx near 0
  2. −∣x∣≤f(x)≤∣x∣-|x| \leq f(x) \leq |x| for all xx near 0 (correct answer)
  3. x2−1≤f(x)≤x2+1x^2 - 1 \leq f(x) \leq x^2 + 1 for all xx near 0
  4. sin⁡x≤f(x)≤cos⁡x\sin x \leq f(x) \leq \cos x for all xx near 0
Explanation: For the Squeeze Theorem to give lim⁡x→0f(x)=0\lim_{x \to 0} f(x) = 0, both bounding functions must approach 0. In choice B, lim⁡x→0(−∣x∣)=0\lim_{x \to 0} (-|x|) = 0 and lim⁡x→0∣x∣=0\lim_{x \to 0} |x| = 0, so the theorem applies. Choice A has bounds approaching different values, choice C has bounds both approaching 1, and choice D has bounds approaching sin⁡(0)=0\sin(0) = 0 and cos⁡(0)=1\cos(0) = 1.

Question 11

What is lim⁡t→0t2sin⁡(3t)t\lim_{t \to 0} \frac{t^2 \sin(3t)}{t} using the Squeeze Theorem?

  1. 00 (correct answer)
  2. 33
  3. 11
  4. The limit does not exist
Explanation: First, simplify: t2sin⁡(3t)t=tsin⁡(3t)\frac{t^2 \sin(3t)}{t} = t \sin(3t) for t≠0t \neq 0. Since −1≤sin⁡(3t)≤1-1 \leq \sin(3t) \leq 1, multiplying by tt gives −∣t∣≤tsin⁡(3t)≤∣t∣-|t| \leq t \sin(3t) \leq |t| (considering the sign of tt). Since lim⁡t→0(−∣t∣)=0\lim_{t \to 0} (-|t|) = 0 and lim⁡t→0∣t∣=0\lim_{t \to 0} |t| = 0, the Squeeze Theorem gives lim⁡t→0tsin⁡(3t)=0\lim_{t \to 0} t \sin(3t) = 0.

Question 12

If 5−x2≤h(x)≤5+x65 - x^2 \leq h(x) \leq 5 + x^6 for all xx near 0, what is lim⁡x→0h(x)\lim_{x \to 0} h(x)?

  1. 00
  2. 55 (correct answer)
  3. 1010
  4. The limit cannot be determined because the bounds have different powers
Explanation: By the Squeeze Theorem, since lim⁡x→0(5−x2)=5−0=5\lim_{x \to 0} (5 - x^2) = 5 - 0 = 5 and lim⁡x→0(5+x6)=5+0=5\lim_{x \to 0} (5 + x^6) = 5 + 0 = 5, and 5−x2≤h(x)≤5+x65 - x^2 \leq h(x) \leq 5 + x^6, we have lim⁡x→0h(x)=5\lim_{x \to 0} h(x) = 5. The fact that the bounds have different powers is irrelevant as long as both approach the same limit.

Question 13

What is lim⁡x→0xcos⁡(1x2)\lim_{x \to 0} x \cos\left(\frac{1}{x^2}\right) using the Squeeze Theorem?

  1. 00 (correct answer)
  2. 11
  3. −1-1
  4. The limit does not exist due to oscillation
Explanation: Since −1≤cos⁡(1x2)≤1-1 \leq \cos\left(\frac{1}{x^2}\right) \leq 1 for all x≠0x \neq 0, we have −∣x∣≤xcos⁡(1x2)≤∣x∣-|x| \leq x \cos\left(\frac{1}{x^2}\right) \leq |x| for all x≠0x \neq 0. Since lim⁡x→0(−∣x∣)=0\lim_{x \to 0} (-|x|) = 0 and lim⁡x→0∣x∣=0\lim_{x \to 0} |x| = 0, the Squeeze Theorem gives lim⁡x→0xcos⁡(1x2)=0\lim_{x \to 0} x \cos\left(\frac{1}{x^2}\right) = 0.

Question 14

If 3x2−x4≤g(x)≤3x2+x43x^2 - x^4 \leq g(x) \leq 3x^2 + x^4 for all xx near 0, what is lim⁡x→0g(x)x2\lim_{x \to 0} \frac{g(x)}{x^2}?

  1. 00
  2. 33 (correct answer)
  3. 11
  4. The limit does not exist
Explanation: Dividing the inequality by x2>0x^2 > 0 (for xx near but not equal to 0), we get 3−x2≤g(x)x2≤3+x23 - x^2 \leq \frac{g(x)}{x^2} \leq 3 + x^2. Since lim⁡x→0(3−x2)=3\lim_{x \to 0} (3 - x^2) = 3 and lim⁡x→0(3+x2)=3\lim_{x \to 0} (3 + x^2) = 3, the Squeeze Theorem gives lim⁡x→0g(x)x2=3\lim_{x \to 0} \frac{g(x)}{x^2} = 3.

Question 15

What is lim⁡x→0x3sin⁡(2x)\lim_{x \to 0} x^3 \sin\left(\frac{2}{x}\right) using the Squeeze Theorem?

  1. 22
  2. 00 (correct answer)
  3. −2-2
  4. The limit does not exist due to the oscillating sine function
Explanation: Since −1≤sin⁡(2x)≤1-1 \leq \sin\left(\frac{2}{x}\right) \leq 1 for all x≠0x \neq 0, multiplying by x3x^3 gives −∣x∣3≤x3sin⁡(2x)≤∣x∣3-|x|^3 \leq x^3 \sin\left(\frac{2}{x}\right) \leq |x|^3 (accounting for the sign of x3x^3). Since lim⁡x→0(−∣x∣3)=0\lim_{x \to 0} (-|x|^3) = 0 and lim⁡x→0∣x∣3=0\lim_{x \to 0} |x|^3 = 0, the Squeeze Theorem gives lim⁡x→0x3sin⁡(2x)=0\lim_{x \to 0} x^3 \sin\left(\frac{2}{x}\right) = 0.

Question 16

If 2−x4≤h(x)≤2+x42 - x^4 \leq h(x) \leq 2 + x^4 for all xx near 0, what is lim⁡x→0h(x)\lim_{x \to 0} h(x)?

  1. 00
  2. 22 (correct answer)
  3. 44
  4. The limit does not exist because the bounds are different
Explanation: By the Squeeze Theorem, since lim⁡x→0(2−x4)=2−0=2\lim_{x \to 0} (2 - x^4) = 2 - 0 = 2 and lim⁡x→0(2+x4)=2+0=2\lim_{x \to 0} (2 + x^4) = 2 + 0 = 2, and 2−x4≤h(x)≤2+x42 - x^4 \leq h(x) \leq 2 + x^4, we have lim⁡x→0h(x)=2\lim_{x \to 0} h(x) = 2. Both bounding functions approach the same limit.