AP Calculus AB Quiz: Determining Limits Using The Squeeze Theorem

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.

Question 1

1. Since $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$, we have $-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$ (correct answer)
2. Since $\sin\left(\frac{1}{x}\right)$ oscillates, the limit does not exist by the Squeeze Theorem
3. Since $0 \leq \sin\left(\frac{1}{x}\right) \leq 1$, we have $0 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$
4. The Squeeze Theorem cannot be applied because $\sin\left(\frac{1}{x}\right)$ is undefined at $x = 0$

Explanation: Since $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$ for all $x \neq 0$, multiplying by $x^2 \geq 0$ gives $-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$. Since both $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} x^2 = 0$, the Squeeze Theorem gives $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$.

Question 2

1. $0$ (correct answer)
2. $1$
3. $-1$
4. The limit does not exist due to oscillation

Explanation: Since $-1 \leq \cos\left(\frac{1}{x^2}\right) \leq 1$ for all $x \neq 0$, we have $-|x| \leq x \cos\left(\frac{1}{x^2}\right) \leq |x|$ for all $x \neq 0$. Since $\lim_{x \to 0} (-|x|) = 0$ and $\lim_{x \to 0} |x| = 0$, the Squeeze Theorem gives $\lim_{x \to 0} x \cos\left(\frac{1}{x^2}\right) = 0$.

Question 3

1. $0$
2. $3$ (correct answer)
3. $1$
4. The limit does not exist

Explanation: Dividing the inequality by $x^2 > 0$ (for $x$ near but not equal to 0), we get $3 - x^2 \leq \frac{g(x)}{x^2} \leq 3 + x^2$. Since $\lim_{x \to 0} (3 - x^2) = 3$ and $\lim_{x \to 0} (3 + x^2) = 3$, the Squeeze Theorem gives $\lim_{x \to 0} \frac{g(x)}{x^2} = 3$.

Question 4

1. $0$
2. $3$ (correct answer)
3. $-3$
4. The limit cannot be determined from this information

Explanation: The inequality $|f(x) - 3| \leq x^2$ is equivalent to $-x^2 \leq f(x) - 3 \leq x^2$, which gives $3 - x^2 \leq f(x) \leq 3 + x^2$. Since $\lim_{x \to 0} (3 - x^2) = 3$ and $\lim_{x \to 0} (3 + x^2) = 3$, the Squeeze Theorem gives $\lim_{x \to 0} f(x) = 3$.

Question 5

1. $2$
2. $0$ (correct answer)
3. $-2$
4. The limit does not exist due to the oscillating sine function

Explanation: Since $-1 \leq \sin\left(\frac{2}{x}\right) \leq 1$ for all $x \neq 0$, multiplying by $x^3$ gives $-|x|^3 \leq x^3 \sin\left(\frac{2}{x}\right) \leq |x|^3$ (accounting for the sign of $x^3$). Since $\lim_{x \to 0} (-|x|^3) = 0$ and $\lim_{x \to 0} |x|^3 = 0$, the Squeeze Theorem gives $\lim_{x \to 0} x^3 \sin\left(\frac{2}{x}\right) = 0$.

Question 6

1. $-x^4 \leq f(x) \leq x^2$ for all $x$ near 0
2. $-|x| \leq f(x) \leq |x|$ for all $x$ near 0 (correct answer)
3. $x^2 - 1 \leq f(x) \leq x^2 + 1$ for all $x$ near 0
4. $\sin x \leq f(x) \leq \cos x$ for all $x$ near 0

Explanation: For the Squeeze Theorem to give $\lim_{x \to 0} f(x) = 0$, both bounding functions must approach 0. In choice B, $\lim_{x \to 0} (-|x|) = 0$ and $\lim_{x \to 0} |x| = 0$, so the theorem applies. Choice A has bounds approaching different values, choice C has bounds both approaching 1, and choice D has bounds approaching $\sin(0) = 0$ and $\cos(0) = 1$.

Question 7

1. $0$ (correct answer)
2. $3$
3. $1$
4. The limit does not exist

Explanation: First, simplify: $\frac{t^2 \sin(3t)}{t} = t \sin(3t)$ for $t \neq 0$. Since $-1 \leq \sin(3t) \leq 1$, multiplying by $t$ gives $-|t| \leq t \sin(3t) \leq |t|$ (considering the sign of $t$). Since $\lim_{t \to 0} (-|t|) = 0$ and $\lim_{t \to 0} |t| = 0$, the Squeeze Theorem gives $\lim_{t \to 0} t \sin(3t) = 0$.

Question 8

1. $0$
2. $5$ (correct answer)
3. $10$
4. The limit cannot be determined because the bounds have different powers

Explanation: By the Squeeze Theorem, since $\lim_{x \to 0} (5 - x^2) = 5 - 0 = 5$ and $\lim_{x \to 0} (5 + x^6) = 5 + 0 = 5$, and $5 - x^2 \leq h(x) \leq 5 + x^6$, we have $\lim_{x \to 0} h(x) = 5$. The fact that the bounds have different powers is irrelevant as long as both approach the same limit.

Question 9

1. $1$
2. $0$ (correct answer)
3. $-1$
4. The limit does not exist due to oscillation

Explanation: Since $-1 \leq \cos\left(\frac{1}{x}\right) \leq 1$ and $\sqrt{x^2 + x^4} \geq 0$, we have $-\sqrt{x^2 + x^4} \leq \sqrt{x^2 + x^4} \cos\left(\frac{1}{x}\right) \leq \sqrt{x^2 + x^4}$. Since $\sqrt{x^2 + x^4} = |x|\sqrt{1 + x^2}$ and $\lim_{x \to 0} |x|\sqrt{1 + x^2} = 0 \cdot 1 = 0$, both bounds approach 0, so by the Squeeze Theorem, the limit is 0.

Question 10

1. $0$
2. $1$ (correct answer)
3. $2$
4. $-1$

Explanation: Since $\lim_{x \to 0} (1 - 2|x|) = 1 - 2(0) = 1$ and $\lim_{x \to 0} (1 + 2|x|) = 1 + 2(0) = 1$, and $1 - 2|x| \leq g(x) \leq 1 + 2|x|$, the Squeeze Theorem gives $\lim_{x \to 0} g(x) = 1$. Both bounding functions approach the same limit of 1.

Question 11

1. $0$
2. $2$ (correct answer)
3. $4$
4. The limit does not exist because the bounds are different

Explanation: By the Squeeze Theorem, since $\lim_{x \to 0} (2 - x^4) = 2 - 0 = 2$ and $\lim_{x \to 0} (2 + x^4) = 2 + 0 = 2$, and $2 - x^4 \leq h(x) \leq 2 + x^4$, we have $\lim_{x \to 0} h(x) = 2$. Both bounding functions approach the same limit.

Question 12

1. The limit equals $0$ by the Squeeze Theorem since both bounds approach $0$
2. The limit equals $1$ by the Squeeze Theorem since both bounds approach $1$
3. The limit cannot be determined by the Squeeze Theorem since the bounds approach different values (correct answer)
4. The limit equals $\frac{1}{2}$ by the Squeeze Theorem as the average of the bounds

Explanation: At $x = \frac{\pi}{2}$, $\cos\left(\frac{\pi}{2}\right) = 0$ and the upper bound is $1$. Since $\lim_{x \to \frac{\pi}{2}} \cos x = 0$ and $\lim_{x \to \frac{\pi}{2}} 1 = 1$, the bounding functions approach different limits ($0$ and $1$). The Squeeze Theorem cannot be applied because both bounds must approach the same value.

Question 13

1. $1$
2. $0$ (correct answer)
3. $-1$
4. The limit oscillates and does not exist

Explanation: Since $-1 \leq \cos\left(\frac{1}{\theta}\right) \leq 1$ for all $\theta \neq 0$, multiplying by $\theta^2 \geq 0$ gives $-\theta^2 \leq \theta^2 \cos\left(\frac{1}{\theta}\right) \leq \theta^2$. Since $\lim_{\theta \to 0} (-\theta^2) = 0$ and $\lim_{\theta \to 0} \theta^2 = 0$, the Squeeze Theorem gives $\lim_{\theta \to 0} \theta^2 \cos\left(\frac{1}{\theta}\right) = 0$.

Question 14

1. $3$
2. $0$ (correct answer)
3. $-3$
4. The limit could be any value between $-3$ and $3$

Explanation: The inequality $|g(x)| \leq 3x^2$ is equivalent to $-3x^2 \leq g(x) \leq 3x^2$. Since $\lim_{x \to 0} (-3x^2) = 0$ and $\lim_{x \to 0} 3x^2 = 0$, the Squeeze Theorem gives $\lim_{x \to 0} g(x) = 0$. The absolute value inequality forces $g(x)$ to be squeezed to 0.

Question 15

1. The functions $g(x)$ and $h(x)$ must be continuous at $x = a$
2. The function $f(x)$ must be defined at $x = a$
3. The limits $\lim_{x \to a} g(x)$ and $\lim_{x \to a} h(x)$ must exist and be equal (correct answer)
4. The inequality $g(x) \leq f(x) \leq h(x)$ must hold for all real numbers

Explanation: The Squeeze Theorem requires that $g(x) \leq f(x) \leq h(x)$ in some neighborhood of $a$ (not necessarily at $a$ itself), and that $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$ for some value $L$. Then $\lim_{x \to a} f(x) = L$. The bounding functions need not be continuous, $f$ need not be defined at $a$, and the inequality need only hold near $a$.

Question 16

1. $0$ (correct answer)
2. $1$
3. $-1$
4. The limit does not exist

Explanation: By the Squeeze Theorem, since $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} x^2 = 0$, and $-x^2 \leq f(x) \leq x^2$, we have $\lim_{x \to 0} f(x) = 0$. The function $f(x)$ is squeezed between two functions that both approach 0.