Determining Limits Using the Squeeze Theorem - AP Calculus AB

1. What is $\lim_{x \to 0} \sqrt{x^2 + x^4} \cos\left(\frac{1}{x}\right)$ using the Squeeze Theorem?
   Answer: $0$ 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.
2. Which condition is essential for applying the Squeeze Theorem to find $\lim_{x \to a} f(x)$?
   Answer: The limits $\lim_{x \to a} g(x)$ and $\lim_{x \to a} h(x)$ must exist and be equal 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 $a$.
3. Which of the following is a correct application of the Squeeze Theorem to find $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$?
   Answer: Since $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$, we have $-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$ 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 $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$.
4. If $-x^2 \leq f(x) \leq x^2$ for all $x$ near 0, what is $\lim_{x \to 0} f(x)$?
   Answer: $0$ 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.
5. Using the Squeeze Theorem, what is $\lim_{\theta \to 0} \theta^2 \cos\left(\frac{1}{\theta}\right)$?
   Answer: $0$ 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 $\lim_{\theta \to 0} \theta^2 \cos\left(\frac{1}{\theta}\right) = 0$.
6. If $1 - 2|x| \leq g(x) \leq 1 + 2|x|$ for all $x$ near 0, what is $\lim_{x \to 0} g(x)$?
   Answer: $1$ 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.
7. What is $\lim_{x \to 0} x \cos\left(\frac{1}{x^2}\right)$ using the Squeeze Theorem?
   Answer: $0$ 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 $\lim_{x \to 0} x \cos\left(\frac{1}{x^2}\right) = 0$.
8. What is $\lim_{t \to 0} \frac{t^2 \sin(3t)}{t}$ using the Squeeze Theorem?
   Answer: $0$ 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 $\lim_{t \to 0} t \sin(3t) = 0$.
9. If $|g(x)| \leq 3x^2$ for all $x$ near 0, what is $\lim_{x \to 0} g(x)$?
   Answer: $0$ 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.
10. If $\cos x \leq f(x) \leq 1$ for all $x$ near $\frac{\pi}{2}$, what can be concluded about $\lim_{x \to \frac{\pi}{2}} f(x)$?
    Answer: The limit cannot be determined by the Squeeze Theorem since the bounds approach different values 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.
11. Which of the following inequalities would allow us to conclude that $\lim_{x \to 0} f(x) = 0$ using the Squeeze Theorem?
    Answer: $-|x| \leq f(x) \leq |x|$ for all $x$ near 0 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 $\cos(0) = 1$.
12. What is $\lim_{x \to 0} x^3 \sin\left(\frac{2}{x}\right)$ using the Squeeze Theorem?
    Answer: $0$ 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 $\lim_{x \to 0} x^3 \sin\left(\frac{2}{x}\right) = 0$.
13. If $3x^2 - x^4 \leq g(x) \leq 3x^2 + x^4$ for all $x$ near 0, what is $\lim_{x \to 0} \frac{g(x)}{x^2}$?
    Answer: $3$ 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 $\lim_{x \to 0} \frac{g(x)}{x^2} = 3$.
14. If $|f(x) - 3| \leq x^2$ for all $x$ near 0, what is $\lim_{x \to 0} f(x)$?
    Answer: $3$ 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 $\lim_{x \to 0} f(x) = 3$.
15. If $5 - x^2 \leq h(x) \leq 5 + x^6$ for all $x$ near 0, what is $\lim_{x \to 0} h(x)$?
    Answer: $5$ 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.
16. If $2 - x^4 \leq h(x) \leq 2 + x^4$ for all $x$ near 0, what is $\lim_{x \to 0} h(x)$?
    Answer: $2$ 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.

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