AP Calculus BC Quiz: Determining Limits Using The Squeeze Theorem
20 questions · exam conditions
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Determining Limits Using The Squeeze TheoremQuestion 1 of 20

If sin⁡xx≤p(x)≤sin⁡xx+x2\dfrac{\sin x}{x}\le p(x)\le \dfrac{\sin x}{x}+x^2 near 00, what is lim⁡x→0p(x)\lim_{x\to 0} p(x)?

00
11
−1-1
lim⁡x→0(sin⁡xx+x2)\lim_{x\to 0}\left(\dfrac{\sin x}{x}+x^2\right)
Does not exist
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AP Calculus BC Quiz

AP Calculus BC Quiz: Determining Limits Using The Squeeze Theorem

Practice Determining Limits Using The Squeeze Theorem in AP Calculus BC 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 BC.

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

If sin⁡xx≤p(x)≤sin⁡xx+x2\dfrac{\sin x}{x}\le p(x)\le \dfrac{\sin x}{x}+x^2 near 00, what is lim⁡x→0p(x)\lim_{x\to 0} p(x)?

  1. 00
  2. 11 (correct answer)
  3. −1-1
  4. lim⁡x→0(sin⁡xx+x2)\lim_{x\to 0}\left(\dfrac{\sin x}{x}+x^2\right)
  5. Does not exist

Explanation: This problem utilizes the squeeze theorem to find the limit of p(x) as x approaches 0. The function is bounded below by (sin x)/x and above by (sin x)/x + x² near 0. As x approaches 0, both bounds approach 1. Therefore, by the squeeze theorem, p(x) must also approach 1. A tempting distractor is A, 0, perhaps if one confuses it with lim x² alone. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.

Question 2

If arctan⁡xx≤S(x)≤1\dfrac{\arctan x}{x}\le S(x)\le 1 near 00, what is lim⁡x→0S(x)\lim_{x\to 0} S(x)?

  1. 00
  2. 11 (correct answer)
  3. π2\dfrac{\pi}{2}
  4. lim⁡x→0arctan⁡xx\lim_{x\to 0}\dfrac{\arctan x}{x}
  5. Does not exist

Explanation: This problem utilizes the squeeze theorem to find the limit of S(x) as x approaches 0. The function is bounded below by arctan⁡xx\dfrac{\arctan x}{x} and above by 1 near 0. As x approaches 0, arctan⁡xx\dfrac{\arctan x}{x} approaches 1 and 1 approaches 1. Therefore, by the squeeze theorem, S(x) must also approach 1. A tempting distractor is A, 0, perhaps if one confuses the limit with that of arctan⁡x\arctan x itself. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.

Question 3

Suppose 1x−3x2≤j(x)≤1x+3x2\dfrac{1}{x}-\dfrac{3}{x^2}\le j(x)\le \dfrac{1}{x}+\dfrac{3}{x^2} for large xx; find lim⁡x→∞x j(x)\lim_{x\to\infty} x\,j(x).

  1. 00
  2. 11 (correct answer)
  3. 33
  4. ∞\infty
  5. −∞-\infty

Explanation: This problem utilizes the squeeze theorem to find the limit of x j(x) as x approaches infinity. The function j(x) is bounded below by 1/x - 3/x² and above by 1/x + 3/x² for large x, so x j(x) is bounded by 1 - 3/x and 1 + 3/x. As x approaches infinity, both 1 - 3/x and 1 + 3/x approach 1. Therefore, by the squeeze theorem, x j(x) must also approach 1. A tempting distractor is C, 3, perhaps if one focuses on the coefficient 3 without multiplying by x. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.

Question 4

Given −(x+2)310≤H(x)≤(x+2)310-\dfrac{(x+2)^3}{10}\le H(x)\le \dfrac{(x+2)^3}{10} near x=−2x=-2, find lim⁡x→−2H(x)\lim_{x\to -2} H(x).

  1. −2-2
  2. 00 (correct answer)
  3. 110\dfrac{1}{10}
  4. −110-\dfrac{1}{10}
  5. Does not exist

Explanation: This problem requires the squeeze theorem to determine the limit of H(x) as x approaches -2. The function H(x) is bounded below by -(x+2)^3/10 and above by (x+2)^3/10, both approaching 0 as x approaches -2. Since lim_{x→-2} -(x+2)^3/10 = 0 and lim_{x→-2} (x+2)^3/10 = 0, the squeeze theorem implies lim_{x→-2} H(x) = 0. The cubic terms vanish at the point. A tempting distractor is choice E, does not exist, perhaps due to the odd power's sign change. A transferable squeeze recognition strategy is to check if higher-odd-power bounds still converge to zero.

Question 5

For x≠0x\ne0, suppose −∣x∣1+∣x∣≤h(x)≤∣x∣1+∣x∣-\dfrac{|x|}{1+|x|}\le h(x)\le \dfrac{|x|}{1+|x|}. Find lim⁡x→0h(x)\lim_{x\to0} h(x).

  1. 12\dfrac{1}{2}
  2. 00 (correct answer)
  3. 11
  4. DNE\text{DNE}
  5. lim⁡x→0∣x∣1+∣x∣\lim_{x\to0} \dfrac{|x|}{1+|x|}

Explanation: This problem applies the squeeze theorem with the bounds -|x|/(1+|x|) ≤ h(x) ≤ |x|/(1+|x|) for x ≠ 0. To find lim(x→0) h(x), we need to evaluate the limits of both bounds as x → 0. As x → 0, we have |x| → 0, so |x|/(1+|x|) → 0/(1+0) = 0. Similarly, -|x|/(1+|x|) → -0/(1+0) = 0. Since both the upper and lower bounds approach 0, the squeeze theorem tells us that lim(x→0) h(x) = 0. Choice A (1/2) might tempt students who incorrectly think the fraction approaches a non-zero value, but careful evaluation shows both numerator and denominator approach definite values with the numerator going to 0. Remember that in squeeze theorem problems, you must verify that both bounds approach the same limit before concluding the squeezed function has that limit.

Question 6

If cos⁡x≤p(x)≤1\cos x\le p(x)\le 1 for all xx, what is lim⁡x→0p(x)\lim_{x\to0} p(x)?

  1. 00
  2. DNE\text{DNE}
  3. −1-1
  4. 11 (correct answer)
  5. lim⁡x→0cos⁡x\lim_{x\to0} \cos x

Explanation: This problem uses the squeeze theorem with the inequality cos x ≤ p(x) ≤ 1 for all x. To find lim(x→0) p(x), we evaluate the limits of both bounds as x → 0. The upper bound is the constant function 1, so its limit is 1. The lower bound cos x has limit cos(0) = 1 as x → 0. Since both bounds approach 1, the squeeze theorem guarantees that lim(x→0) p(x) = 1. Choice C (-1) might attract students who confuse cos(0) with cos(π), but cos(0) = 1 is a fundamental value to memorize. When applying the squeeze theorem, always evaluate the limits of both bounds carefully—here, both bounds converge to the same value 1, forcing p(x) to have limit 1 as well.

Question 7

Suppose sin⁡2xx2≤Y(x)≤1\dfrac{\sin^2 x}{x^2}\le Y(x)\le 1 for xx near 00, find lim⁡x→0Y(x)\lim_{x\to 0} Y(x).

  1. 00
  2. 11 (correct answer)
  3. −1-1
  4. lim⁡x→0sin⁡2xx2\lim_{x\to 0}\dfrac{\sin^2 x}{x^2}
  5. Does not exist

Explanation: This problem utilizes the squeeze theorem to find the limit of Y(x) as x approaches 0. The function is bounded below by sin²x / x² and above by 1 near 0. As x approaches 0, sin²x / x² approaches 1 and 1 approaches 1. Therefore, by the squeeze theorem, Y(x) must also approach 1. A tempting distractor is A, 0, perhaps if one confuses it with lim sin x / x without squaring. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.

Question 8

If −∣x−1∣2≤z(x)≤∣x−1∣2-\dfrac{|x-1|}{2}\le z(x)\le \dfrac{|x-1|}{2} near x=1x=1, find lim⁡x→1z(x)\lim_{x\to 1} z(x).

  1. 12\dfrac{1}{2}
  2. 00 (correct answer)
  3. −12-\dfrac{1}{2}
  4. 11
  5. Does not exist

Explanation: This problem requires the squeeze theorem to determine the limit of z(x) as x approaches 1. The function z(x) is bounded below by -|x-1|/2 and above by |x-1|/2, both approaching 0 as x approaches 1. Since lim_{x→1} -|x-1|/2 = 0 and lim_{x→1} |x-1|/2 = 0, the squeeze theorem guarantees lim_{x→1} z(x) = 0. The absolute value creates linear convergence to zero. A tempting distractor is choice E, does not exist, maybe due to the absolute value's kink. A transferable squeeze recognition strategy is to employ absolute value bounds for limits at non-zero points.

Question 9

Given −∣x∣37≤g(x)≤∣x∣37-\dfrac{|x|^3}{7} \le g(x) \le \dfrac{|x|^3}{7} near 00, find lim⁡x→0g(x)\lim_{x\to 0} g(x).

  1. 17\dfrac{1}{7}
  2. 00 (correct answer)
  3. −17-\dfrac{1}{7}
  4. ∞\infty
  5. Does not exist

Explanation: This problem utilizes the squeeze theorem to find the limit of g(x) as x approaches 0. The function is bounded below by −∣x∣37-\frac{|x|^3}{7} and above by ∣x∣37\frac{|x|^3}{7} near 0. As x approaches 0, both bounds approach 0. Therefore, by the squeeze theorem, g(x) must also approach 0. A tempting distractor is A, 17\frac{1}{7}, perhaps if one ignores the ∣x∣3|x|^3 term. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.

Question 10

Given 2x2≤V(x)≤5x22x^2\le V(x)\le 5x^2 near 00, what is lim⁡x→0V(x)\lim_{x\to 0} V(x)?

  1. 22
  2. 55
  3. 00 (correct answer)
  4. lim⁡x→0(2x2)\lim_{x\to 0}(2x^2)
  5. Does not exist

Explanation: This problem utilizes the squeeze theorem to find the limit of V(x) as x approaches 0. The function is bounded below by 2x² and above by 5x² near 0. As x approaches 0, both 2x² and 5x² approach 0. Therefore, by the squeeze theorem, V(x) must also approach 0. A tempting distractor is B, 5, perhaps if one mistakenly takes the coefficient of the upper bound as the limit. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.

Question 11

If 1x+1−1x2≤U(x)≤1x+1+1x2\dfrac{1}{x+1}-\dfrac{1}{x^2}\le U(x)\le \dfrac{1}{x+1}+\dfrac{1}{x^2} for large xx, find lim⁡x→∞U(x)\lim_{x\to\infty} U(x).

  1. 11
  2. 00 (correct answer)
  3. −1-1
  4. lim⁡x→∞1x+1\lim_{x\to\infty}\dfrac{1}{x+1}
  5. ∞\infty

Explanation: This problem utilizes the squeeze theorem to find the limit of U(x) as x→∞x \to \infty. The function is bounded below by 1x+1−1x2\frac{1}{x+1} - \frac{1}{x^2} and above by 1x+1+1x2\frac{1}{x+1} + \frac{1}{x^2} for large xx. As x→∞x \to \infty, both bounds approach 00. Therefore, by the squeeze theorem, U(x) must also approach 00. A tempting distractor is A, 11, perhaps if one ignores the added terms and focuses only on 1x+1\frac{1}{x+1}. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.

Question 12

Suppose −∣x∣≤T(x)≤∣x∣-\sqrt{|x|}\le T(x)\le \sqrt{|x|} near 00; find lim⁡x→0T(x)\lim_{x\to 0} T(x).

  1. 11
  2. 00 (correct answer)
  3. −1-1
  4. lim⁡x→0∣x∣\lim_{x\to 0}\sqrt{|x|}
  5. Does not exist

Explanation: This problem utilizes the squeeze theorem to find the limit of T(x) as x approaches 0. The function is bounded below by -√|x| and above by √|x| near 0. As x approaches 0, both -√|x| and √|x| approach 0. Therefore, by the squeeze theorem, T(x) must also approach 0. A tempting distractor is E, 'Does not exist,' perhaps if one thinks the absolute value causes oscillation, but the bounds converge. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.

Question 13

If −x≤g(x)≤x-\sqrt{x}\le g(x)\le \sqrt{x} for x≥0x\ge0, what is lim⁡x→0+g(x)\lim_{x\to0^+} g(x)?

  1. DNE\text{DNE}
  2. 11
  3. 00 (correct answer)
  4. lim⁡x→0+x\lim_{x\to0^+} \sqrt{x}
  5. −1-1

Explanation: This problem uses the squeeze theorem to find the right-hand limit of g(x) as x approaches 0 from the positive side. We have -√x ≤ g(x) ≤ √x for x ≥ 0. As x → 0⁺, both bounds approach the same value: -√x → 0 and √x → 0. Since g(x) is squeezed between two functions that both approach 0, the squeeze theorem guarantees that lim(x→0⁺) g(x) = 0. Choice A (DNE) might appeal to students who worry about the square root being undefined at x = 0, but we're taking a limit as x approaches 0, not evaluating at 0. The key insight for squeeze theorem problems is to check whether both bounding functions approach the same limit—if they do, the squeezed function must approach that same value.

Question 14

If −∣x∣x2+4≤Z(x)≤∣x∣x2+4-\dfrac{|x|}{x^2+4}\le Z(x)\le \dfrac{|x|}{x^2+4} near 00, what is lim⁡x→0Z(x)\lim_{x\to 0} Z(x)?

  1. 14\dfrac{1}{4}
  2. 00 (correct answer)
  3. −14-\dfrac{1}{4}
  4. lim⁡x→0∣x∣x2+4\lim_{x\to 0}\dfrac{|x|}{x^2+4}
  5. Does not exist

Explanation: This problem utilizes the squeeze theorem to find the limit of Z(x) as x approaches 0. The function is bounded below by -|x|/(x²+4) and above by |x|/(x²+4) near 0. As x approaches 0, both bounds approach 0. Therefore, by the squeeze theorem, Z(x) must also approach 0. A tempting distractor is A, 1/4, perhaps if one plugs in x=0 incorrectly into the denominator. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.

Question 15

If −∣x−5∣x2+1≤r(x)≤∣x−5∣x2+1-\dfrac{|x-5|}{x^2+1} \le r(x) \le \dfrac{|x-5|}{x^2+1} near x=5x=5, what is lim⁡x→5r(x)\lim_{x\to 5} r(x)?

  1. 126\dfrac{1}{26}
  2. 00 (correct answer)
  3. −126-\dfrac{1}{26}
  4. 55
  5. Does not exist

Explanation: This problem utilizes the squeeze theorem to find the limit of r(x) as x approaches 5. The function is bounded below by −∣x−5∣x2+1-\frac{|x-5|}{x^2 + 1} and above by ∣x−5∣x2+1\frac{|x-5|}{x^2 + 1} near x=5. As x approaches 5, both bounds approach 0. Therefore, by the squeeze theorem, r(x) must also approach 0. A tempting distractor is D, 5, perhaps if one focuses on the point x=5 without the limit. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.

Question 16

Given 7−1n≤en≤7+1n7-\dfrac{1}{\sqrt{n}}\le e_n\le 7+\dfrac{1}{\sqrt{n}}, what is lim⁡n→∞en\lim_{n\to\infty} e_n?

  1. 00
  2. 77 (correct answer)
  3. 11
  4. ∞\infty
  5. Does not exist

Explanation: This problem utilizes the squeeze theorem to find the limit of the sequence ene_n as nn approaches infinity. The sequence is bounded below by 7−1n7 - \frac{1}{\sqrt{n}} and above by 7+1n7 + \frac{1}{\sqrt{n}}. As nn approaches infinity, both bounds approach 77. Therefore, by the squeeze theorem, ene_n must also approach 77. A tempting distractor is A, 00, perhaps if one ignores the constant 77. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.

Question 17

Suppose −1x≤e(x)≤1x-\dfrac{1}{x}\le e(x)\le \dfrac{1}{x} for xx near ∞\infty; find lim⁡x→∞e(x)\lim_{x\to\infty} e(x).

  1. 11
  2. 00 (correct answer)
  3. −1-1
  4. ∞\infty
  5. Does not exist

Explanation: This problem utilizes the squeeze theorem to find the limit of e(x) as x approaches infinity. The function is bounded below by -1/x and above by 1/x for large positive x. As x approaches infinity, both -1/x and 1/x approach 0. Therefore, by the squeeze theorem, e(x) must also approach 0. A tempting distractor is D, ∞, perhaps if one thinks the 1/x term diverges, but it approaches 0. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.

Question 18

If −1x≤w(x)≤1x-\dfrac{1}{\sqrt{x}}\le w(x)\le \dfrac{1}{\sqrt{x}} for xx near ∞\infty, find lim⁡x→∞w(x)\lim_{x\to\infty} w(x).

  1. ∞\infty
  2. 00 (correct answer)
  3. 11
  4. −1-1
  5. Does not exist

Explanation: This problem requires the squeeze theorem to determine the limit of w(x) as x approaches infinity. The function w(x) is bounded below by -1/√x and above by 1/√x, both approaching 0 as x approaches infinity. Since lim_{x→∞} -1/√x = 0 and lim_{x→∞} 1/√x = 0, the squeeze theorem guarantees lim_{x→∞} w(x) = 0. The square root in the denominator drives the terms to zero. A tempting distractor is choice A, ∞, maybe thinking square roots grow indefinitely. A transferable squeeze recognition strategy is to use reciprocal roots for limits at infinity converging to zero.

Question 19

If 0≤k(x)≤(x+1)21+(x+1)20\le k(x)\le \dfrac{(x+1)^2}{1+(x+1)^2} near x=−1x=-1, what is lim⁡x→−1k(x)\lim_{x\to -1} k(x)?

  1. 11
  2. 00 (correct answer)
  3. −1-1
  4. lim⁡x→−1(x+1)21+(x+1)2\lim_{x\to -1}\dfrac{(x+1)^2}{1+(x+1)^2}
  5. Does not exist

Explanation: This problem utilizes the squeeze theorem to find the limit of k(x) as x approaches -1. The function is bounded below by 0 and above by (x+1)² / (1 + (x+1)²) near x=-1. As x approaches -1, both 0 and the upper bound approach 0. Therefore, by the squeeze theorem, k(x) must also approach 0. A tempting distractor is A, 1, perhaps if one thinks the denominator approaches 1 without the numerator. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.

Question 20

Given 4−2n2≤cn≤4+2n24-\dfrac{2}{n^2}\le c_n\le 4+\dfrac{2}{n^2} for n≥1n\ge 1, find lim⁡n→∞cn\lim_{n\to\infty} c_n.

  1. 22
  2. 44 (correct answer)
  3. 00
  4. lim⁡n→∞(4+2n2)\lim_{n\to\infty}\left(4+\dfrac{2}{n^2}\right)
  5. Does not exist

Explanation: This problem utilizes the squeeze theorem to find the limit of the sequence c_n as n approaches infinity. The sequence is bounded below by 4−2n24 - \dfrac{2}{n^2} and above by 4+2n24 + \dfrac{2}{n^2} for n≥1n \ge 1. As n approaches infinity, both bounds approach 44. Therefore, by the squeeze theorem, c_n must also approach 44. A tempting distractor is A, 2, perhaps if one focuses only on the added term. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.