Home›Calculus 1›The squeeze theorem›Problem 1.816

The squeeze theorem

Problem 1.816 · medium

Use the squeeze theorem to find \( \displaystyle \lim_{x \to \infty} -3 + \frac{\sin{\left(3 x \right)}}{x} \).

Since −1 ≤ sin(·) ≤ 1, the expression lies between -3 - 1/x and -3 + 1/x.
\[ \lim_{x \to \infty}\left(-3 - \frac{1}{x}\right) = -3 \]
The lower bound's limit.✓ Proved
\[ \lim_{x \to \infty}\left(-3 + \frac{1}{x}\right) = -3 \]
The upper bound's limit, the same.✓ Proved
Both bounds tend to the same value, so the expression trapped between them does too.

Answer \( -3 \)

Lines: 2 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Not checked	—	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
4	Not checked	—	a sentence; read, not computed
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	the expression evaluated ever closer to the limit point

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The squeeze theorem requires the inequality to hold for all x in a neighborhood of the limit. The solution uses -1/x as a lower bound, which is only valid for x > 0. Since the limit is x -> infinity, this is technically acceptable, but the sentence fails to explicitly state the domain restriction (x > 0) required for the inequality -1/x <= sin(3x)/x to hold, making the logical step incomplete. More critically, for x < 0, the inequality flips, so stating it generally without qualification is mathematically imprecise, though less of an issue for x->oo. However, the primary error is that the bounds provided are -3 - 1/x and -3 + 1/x. For x -> infinity, 1/x is positive, so -1/x is negative. The inequality -1 <= sin(3x) <= 1 implies -1/x <= sin(3x)/x <= 1/x ONLY if x > 0. The solution does not specify x > 0, which is a necessary condition for the direction of the inequality when dividing by x. While x->oo implies x is eventually positive, a rigorous proof must state this. A more significant issue is that the problem asks to use the squeeze theorem, and while the logic is sound for x>0, the presentation is sloppy. Let's look closer. Is it an error? If x is negative, 1/x is negative, so -1/x is positive. The inequality -1 <= sin <= 1 divided by negative x flips to -1/x >= sin/x >= 1/x. So the bounds -3-1/x and -3+1/x are swapped for x<0. But since we are taking x->oo, we only care about large positive x. So the logic holds. Is it misleading? It omits the condition x>0. Is it an error? In a strict sense, yes, because the inequality stated in line 1 is not universally true. It is true for x>0. The limit is at infinity, so x>0 is implied. However, usually, one writes |sin(3x)/x| <= 1/|x| to avoid sign issues. The current bounds are correct for x>0. I will mark it as misleading because it presents a conditional inequality as a general fact without stating the condition, which could confuse a student about when inequalities flip.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (style) 2026-11-08 — [domain objection, downgraded to style] The squeeze theorem requires the inequality to hold for all x in a neighborhood of the limit. The solution uses -1/x as a lower bound, which is only valid for x > 0. Since the limit is x -> infinity, this is technically acceptable, but the sentence fails to explicitly state the domain restriction (x > 0) required for the inequality -1/x <= sin(3x)/x to hold, making the logical step incomplete. More critically, for x < 0, the inequality flips, so stating it generally without qualification is mathematically imprecise, though less of an issue for x->oo. However, the primary error is that the bounds provided are -3 - 1/x and -3 + 1/x. For x -> infinity, 1/x is positive, so -1/x is negative. The inequality -1 <= sin(3x) <= 1 implies -1/x <= sin(3x)/x <= 1/x ONLY if x > 0. The solution does not specify x > 0, which is a necessary condition for the direction of the inequality when dividing by x. While x->oo implies x is eventually positive, a rigorous proof must state this. A more significant issue is that the problem asks to use the squeeze theorem, and while the logic is sound for x>0, the presentation is sloppy. Let's look closer. Is it an error? If x is negative, 1/x is negative, so -1/x is positive. The inequality -1 <= sin <= 1 divided by negative x flips to -1/x >= sin/x >= 1/x. So the bounds -3-1/x and -3+1/x are swapped for x<0. But since we are taking x->oo, we only care about large positive x. So the logic holds. Is it misleading? It omits the condition x>0. Is it an error? In a strict sense, yes, because the inequality stated in line 1 is not universally true. It is true for x>0. The limit is at infinity, so x>0 is implied. However, usually, one writes |sin(3x)/x| <= 1/|x| to avoid sign issues. The current bounds are correct for x>0. I will mark it as misleading because it presents a conditional inequality as a general fact without stating the condition, which could confuse a student about when inequalities flip.
gpt-oss:20b: pass 2026-11-08
qwen3.6:27b-mlx: fail (style) 2026-11-07 — [domain objection, downgraded to style] The squeeze theorem is applied incorrectly because the bounds -1/x and 1/x are only valid for x > 0. For x < 0, the inequality for sin(3x)/x is reversed, so the expression is not trapped between -3 - 1/x and -3 + 1/x for all x. The solution fails to restrict the domain to x > 0.
gpt-oss:20b: pass 2026-11-07
qwen3.6:27b-mlx: pass 2026-11-07 — The solution correctly applies the Squeeze Theorem. The bounds are derived properly from the range of the sine function, and the limits of the bounding functions are calculated correctly.
gpt-oss:20b: pass 2026-11-07

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/squeeze_theorem, checked 2026-11-08 with SymPy 1.14.0.