Determining Limits Using the Squeeze Theorem - AP Calculus BC

1. Given $-\dfrac{2}{n}\le d_n\le \dfrac{2}{n}$ for $n\ge 1$, find $\lim_{n\to\infty} d_n$.
   Answer: $0$ This problem utilizes the squeeze theorem to find the limit of the sequence $d_n$ as n approaches infinity. The sequence is bounded below by -2/n and above by 2/n for n ≥ 1. As n approaches infinity, both -2/n and 2/n approach 0. Therefore, by the squeeze theorem, $d_n$ must also approach 0. A tempting distractor is A, 2, perhaps if one takes the numerator 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.
2. If $0\le k(x)\le \dfrac{(x+1)^2}{1+(x+1)^2}$ near $x=-1$, what is $\lim_{x\to -1} k(x)$?
   Answer: $0$ 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.
3. For $x\ne0$, $-x^4\le q(x)\le x^4$. What is $\lim_{x\to0} \dfrac{q(x)}{x^2}$?
   Answer: $0$ This problem requires the squeeze theorem to find lim(x→0) q(x)/x². Given -x⁴ ≤ q(x) ≤ x⁴ for x ≠ 0, we divide all parts by x² (valid since x ≠ 0) to get -x² ≤ q(x)/x² ≤ x². As x → 0, both bounds approach the same value: -x² → 0 and x² → 0. By the squeeze theorem, lim(x→0) q(x)/x² = 0. Choice B (1) might tempt students who think x⁴/x² → 1, but x⁴/x² = x² → 0 as x → 0. The key technique here is to manipulate the given inequality to match the desired expression before applying the squeeze theorem—dividing by x² transforms the bounds into forms whose limits are easy to evaluate.
4. Given $-\dfrac{(x+2)^3}{10}\le H(x)\le \dfrac{(x+2)^3}{10}$ near $x=-2$, find $\lim_{x\to -2} H(x)$.
   Answer: $0$ 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.
5. If $-\sqrt{x}\le g(x)\le \sqrt{x}$ for $x\ge0$, what is $\lim_{x\to0^+} g(x)$?
   Answer: $0$ 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.
6. If $\dfrac{2x}{1+x^2}\le v(x)\le \dfrac{2x}{1+x^2}+x^3$, find $\lim_{x\to0} v(x)$.
   Answer: $0$ This problem applies the squeeze theorem with $\frac{2x}{1+x^2} \leq v(x) \leq \frac{2x}{1+x^2} + x^3$. To find $\lim_{x\to0} v(x)$, we evaluate the limits of both bounds. For the lower bound, as $x \to 0$, we get $\frac{2(0)}{1+0^2} = \frac{0}{1} = 0$. For the upper bound, we have $\lim_{x\to0} \left[ \frac{2x}{1+x^2} + x^3 \right] = 0 + 0 = 0$. Since both bounds approach 0, the squeeze theorem gives $\lim_{x\to0} v(x) = 0$. Choice A (1) might attract students who incorrectly evaluate $\frac{2x}{1+x^2}$ at x = 0, but careful substitution shows this equals 0. The key insight is that when bounds differ by a term that vanishes at the limit point (here $x^3 \to 0$), they squeeze the function to the same limit value.
7. Suppose $\dfrac{\sin^2 x}{x^2}\le Y(x)\le 1$ for $x$ near $0$, find $\lim_{x\to 0} Y(x)$.
   Answer: $1$ 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.
8. Suppose $-\dfrac{1}{x}\le e(x)\le \dfrac{1}{x}$ for $x$ near $\infty$; find $\lim_{x\to\infty} e(x)$.
   Answer: $0$ 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.
9. Given $4-\dfrac{2}{n^2}\le c_n\le 4+\dfrac{2}{n^2}$ for $n\ge 1$, find $\lim_{n\to\infty} c_n$.
   Answer: $4$ 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 $4 - \dfrac{2}{n^2}$ and above by $4 + \dfrac{2}{n^2}$ for $n \ge 1$. As n approaches infinity, both bounds approach $4$. Therefore, by the squeeze theorem, $c_n$ must also approach $4$. 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.
10. Given $7-\dfrac{1}{\sqrt{n}}\le e_n\le 7+\dfrac{1}{\sqrt{n}}$, what is $\lim_{n\to\infty} e_n$?
    Answer: $7$ This problem utilizes the squeeze theorem to find the limit of the sequence $e_n$ as $n$ approaches infinity. The sequence is bounded below by $7 - \frac{1}{\sqrt{n}}$ and above by $7 + \frac{1}{\sqrt{n}}$. As $n$ approaches infinity, both bounds approach $7$. Therefore, by the squeeze theorem, $e_n$ must also approach $7$. A tempting distractor is A, $0$, perhaps if one ignores the constant $7$. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.
11. Assume $|u(x)-3| \le x^2$ for all $x$. What is $\lim_{x \to 0} u(x)$?
    Answer: $3$ This problem uses the squeeze theorem with the condition $|u(x) - 3| \leq x^2$. This absolute value inequality is equivalent to $-x^2 \leq u(x) - 3 \leq x^2$, which gives us $3 - x^2 \leq u(x) \leq 3 + x^2$. As $x \to 0$, the lower bound approaches $3 - 0 = 3$ and the upper bound approaches $3 + 0 = 3$. Since both bounds approach 3, the squeeze theorem tells us that $\lim_{x \to 0} u(x) = 3$. Choice A (0) might tempt students who focus on $x^2 \to 0$ without considering the full inequality, but the bounds squeeze u(x) toward 3, not 0. When you see $|f(x) - L| \leq g(x)$ where $g(x) \to 0$, this is the squeeze theorem in disguise—it forces f(x) $\to$ L.
12. For all $x\ne0$, $-\dfrac{1}{1+x^2}\le t(x)\le \dfrac{1}{1+x^2}$. Find $\lim_{x\to\infty} t(x)$.
    Answer: $0$ This problem uses the squeeze theorem with -1/(1+x²) ≤ t(x) ≤ 1/(1+x²) for all x ≠ 0. To find lim(x→∞) t(x), we evaluate the limits of both bounds as x → ∞. As x → ∞, we have 1+x² → ∞, so 1/(1+x²) → 0. Similarly, -1/(1+x²) → 0. Since both bounds approach 0, the squeeze theorem guarantees that lim(x→∞) t(x) = 0. Choice A (1) might attract students who forget that the denominator grows without bound, but 1/(1+x²) → 0, not 1, as x → ∞. When applying the squeeze theorem at infinity, carefully evaluate how rational functions behave—here the quadratic term in the denominator dominates, driving both bounds to 0.
13. Given $-\dfrac{|x|^3}{7} \le g(x) \le \dfrac{|x|^3}{7}$ near $0$, find $\lim_{x\to 0} g(x)$.
    Answer: $0$ This problem utilizes the squeeze theorem to find the limit of g(x) as x approaches 0. The function is bounded below by $-\frac{|x|^3}{7}$ and above by $\frac{|x|^3}{7}$ 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, $\frac{1}{7}$, perhaps if one ignores the $|x|^3$ 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.
14. If $\dfrac{\arctan x}{x}\le S(x)\le 1$ near $0$, what is $\lim_{x\to 0} S(x)$?
    Answer: $1$ This problem utilizes the squeeze theorem to find the limit of S(x) as x approaches 0. The function is bounded below by $\dfrac{\arctan x}{x}$ and above by 1 near 0. As x approaches 0, $\dfrac{\arctan x}{x}$ 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 $\arctan x$ 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.
15. If $-\dfrac{|x-5|}{x^2+1} \le r(x) \le \dfrac{|x-5|}{x^2+1}$ near $x=5$, what is $\lim_{x\to 5} r(x)$?
    Answer: $0$ This problem utilizes the squeeze theorem to find the limit of r(x) as x approaches 5. The function is bounded below by $-\frac{|x-5|}{x^2 + 1}$ and above by $\frac{|x-5|}{x^2 + 1}$ 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.
16. If $\dfrac{1}{x+1}-\dfrac{1}{x^2}\le U(x)\le \dfrac{1}{x+1}+\dfrac{1}{x^2}$ for large $x$, find $\lim_{x\to\infty} U(x)$.
    Answer: $0$ This problem utilizes the squeeze theorem to find the limit of U(x) as $x \to \infty$. The function is bounded below by $\frac{1}{x+1} - \frac{1}{x^2}$ and above by $\frac{1}{x+1} + \frac{1}{x^2}$ for large $x$. As $x \to \infty$, both bounds approach $0$. Therefore, by the squeeze theorem, U(x) must also approach $0$. A tempting distractor is A, $1$, perhaps if one ignores the added terms and focuses only on $\frac{1}{x+1}$. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.
17. If $-\dfrac{1}{\sqrt{x}}\le w(x)\le \dfrac{1}{\sqrt{x}}$ for $x$ near $\infty$, find $\lim_{x\to\infty} w(x)$.
    Answer: $0$ 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.
18. Suppose $-\sqrt{|x|}\le T(x)\le \sqrt{|x|}$ near $0$; find $\lim_{x\to 0} T(x)$.
    Answer: $0$ 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.
19. Given $2x^2\le V(x)\le 5x^2$ near $0$, what is $\lim_{x\to 0} V(x)$?
    Answer: $0$ 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.
20. If $\dfrac{x^2}{2}\le m(x)\le \dfrac{x^2}{2}+|x|$ near $0$, what is $\lim_{x\to 0} m(x)$?
    Answer: $0$ This problem utilizes the squeeze theorem to find the limit of m(x) as x approaches 0. The function is bounded below by x²/2 and above by x²/2 + |x| near 0. As x approaches 0, both bounds approach 0. Therefore, by the squeeze theorem, m(x) must also approach 0. A tempting distractor is A, 1/2, perhaps if one ignores the x terms. A transferable squeeze recognition strategy is to identify bounds that both converge to the same value, forcing the sandwiched term to do the same.

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