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.

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.

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.

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.

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.

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.

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.

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.

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.

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.