Determining Limits Using the Squeeze Theorem: Practice Questions (Calculus AB 1.8)

Answer: (B). The inequality holds for every x, so it holds on an open interval around −2. The lower bound tends to −4(−2) − 4 = 4 and the upper bound tends to (−2)² = 4. The outer limits are equal, so by the squeeze theorem the limit of f is 4. (The two bounds are consistent: x² − (−4x − 4) = (x + 2)², which is never negative.)

• (A) is the limit of the gap between the bounds, (x + 2)², not of f. The gap shrinking to 0 is why the squeeze works, but it is not the answer.
• (C) adds the two outer limits, 4 + 4. The squeeze gives their common value, not their sum.
• (D) misses the point of the theorem: it finds a limit from bounds alone, without a formula.

Answer: (D). Rewrite: (sin 4x)/(3x) = (4/3) × (sin 4x)/(4x). As x → 0, u = 4x → 0, so (sin 4x)/(4x) → 1. The limit is 4/3 × 1 = 4/3.

• (A) treats 0/0 as 0.
• (B) turns the fraction upside down when inserting the 4: it writes (3/4) × (sin 4x)/(4x) instead of (4/3).
• (C) assumes “sin of something over something” always tends to 1. That needs the same quantity inside the sine and in the denominator, and here they are 4x and 3x.

Answer: (B). For x ≠ 0, |x cos(2/x)| = |x| × |cos(2/x)| ≤ |x|, which gives (B) on both sides of 0. Both bounds tend to 0, so the limit is 0.

• (A) fails for x < 0. At x = −0.1 it would say 0.1 ≤ x cos(2/x) ≤ −0.1, which is impossible, since the “floor” is above the “ceiling”. Multiplying by a negative x flips the signs.
• (C) is not true for all x ≠ 0: at x = 10, x cos(2/x) = 10 cos 0.2 ≈ 9.80. It does hold near 0 (when |x| ≤ 1), but even there the bounds tend to −1 and 1. The outer limits differ, so the theorem gives no conclusion.
• (D) is false. At x = 0.1, x cos(2/x) = 0.1 cos 20 ≈ 0.0408, which is bigger than x² = 0.01. Bounds that shrink faster than |x| cannot trap a function of size up to |x|.

Answer: (B). Split the fraction: (1/2) × [(sin x)/x + (1 − cos x)/x]. As x → 0, (sin x)/x → 1 and (1 − cos x)/x → 0. So the limit is (1/2)(1 + 0) = 1/2.

• (A) treats 0/0 as 0.
• (C) mixes up the two key limits and uses 1 for (1 − cos x)/x too: (1/2)(1 + 1) = 1.
• (D) assumes a 0 in the denominator always means no limit. Here the top is also 0, so the form is indeterminate, and the known trig limits settle it.

(a) The inequality holds on the open interval (3, 7), which contains 4, except at x = 4 itself. That is enough. The lower bound tends to 5 − 0 = 5 and the upper bound tends to 5 + 0 = 5. The outer limits are equal, so by the squeeze theorem lim (x → 4) f(x) = 5.

(b) Nothing. The inequality is not claimed at x = 4, so f(4) may be any number, or f may be undefined there. The graph near x = 4 is pinched towards the point (4, 5), whether or not that point is on the graph.

(c) Near x = 6 the graph lies between the downward parabola y = 5 − (x − 4)², which passes through (6, 1), and the V-shaped line y = 5 + |x − 4|, which passes through (6, 7). The bounds tend to 1 and 7. They are different, so the squeeze theorem gives no value for lim (x → 6) f(x). The limit might exist, or might not.

Suggested mark points (3): 1 for the limit 5 with all three conditions named (inequality, open interval around 4 except 4, equal outer limits); 1 for stating that f(4) cannot be determined, with the reason; 1 for bounds 1 and 7 at x = 6 and the conclusion that the theorem gives no answer (not “the limit does not exist”).

(a) The product law needs both lim (x → 3) (x − 3) and lim (x → 3) cos(1/(x − 3)) to exist. The second does not: as x → 3, 1/(x − 3) grows without bound in size, so the cosine keeps swinging between −1 and 1.

(b) For x ≠ 3, |(x − 3) cos(1/(x − 3))| = |x − 3| × |cos(1/(x − 3))| ≤ |x − 3|. So

2 − |x − 3| ≤ g(x) ≤ 2 + |x − 3|.

An absolute value is needed because x − 3 is negative when x < 3. Multiplying −1 ≤ cos(…) ≤ 1 by a negative number reverses the signs, so 2 − (x − 3) ≤ g(x) ≤ 2 + (x − 3) is false to the left of 3. For example, at x = 2.9 it would claim 2.1 ≤ g(2.9) ≤ 1.9.

(c) The inequality holds for all x ≠ 3, so on an open interval around 3 except at 3. lim (x → 3) (2 − |x − 3|) = 2 and lim (x → 3) (2 + |x − 3|) = 2. The outer limits are equal, so by the squeeze theorem lim (x → 3) g(x) = 2.

(d) The student assumes that if one factor has no limit, the product has none. The product law only works in one direction: if both limits exist, so does the limit of the product. It says nothing when one fails. Here the factor (x − 3) shrinks to 0 while cos(1/(x − 3)) stays between −1 and 1, so the product is squeezed to 0 and g tends to 2.

Acceptable alternative for (b): split into cases, 2 + (x − 3) ≤ g(x) ≤ 2 − (x − 3) for x < 3 and 2 − (x − 3) ≤ g(x) ≤ 2 + (x − 3) for x > 3, then apply the theorem to each one-sided limit.

(a) For 0 < |x| < π/2, cos x > 0, and (tan x)/x = [(sin x)/x] ÷ cos x. Divide every part of the given inequality by the positive number cos x:

cos x / cos x ≤ (sin x)/(x cos x) ≤ 1/cos x, which is 1 ≤ (tan x)/x ≤ 1/cos x.

(b) The inequality holds for all x with 0 < |x| < π/2, an open interval around 0 with 0 removed. lim (x → 0) 1 = 1 and lim (x → 0) 1/cos x = 1/cos 0 = 1. The outer limits are equal, so by the squeeze theorem lim (x → 0) (tan x)/x = 1.

(c) Write (tan 3x)/(5x) = (3/5) × (tan 3x)/(3x). As x → 0, u = 3x → 0, so (tan 3x)/(3x) → 1 by (b). The limit is 3/5.

(d) At x = 0.01, (tan 0.03)/0.05 ≈ 0.030009/0.05 ≈ 0.6002. That is very close to 3/5 = 0.6, which supports the answer.

Acceptable alternative for (b): (tan x)/x = [(sin x)/x] × [1/cos x], then use lim (sin x)/x = 1 and the product law. This earns the (b) point only if the result for (sin x)/x is quoted correctly; part (a) still needs the inequality.