The squeeze theorem
Problem 1.813 · medium
Use the squeeze theorem to find \( \displaystyle \lim_{x \to \infty} 2 + \frac{\cos{\left(4 x \right)}}{x} \).
- Since −1 ≤ cos(·) ≤ 1, the expression lies between 2 - 1/x and 2 + 1/x.
- \[ \lim_{x \to \infty}\left(2 - \frac{1}{x}\right) = 2 \]The lower bound's limit.✓ Proved
- \[ \lim_{x \to \infty}\left(2 + \frac{1}{x}\right) = 2 \]The upper bound's limit, the same.✓ Proved
- Both bounds tend to the same value, so the expression trapped between them does too.
Answer \( 2 \)
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: passqwen3.6:27b-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 point. For x < 0, dividing by x reverses the inequality, so the bounds are swapped. The solution fails to restrict the domain to x > 0, making the setup technically incorrect for the general case.
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 point. For x < 0, dividing by x reverses the inequality, so the bounds are swapped. The solution fails to restrict the domain to x > 0, making the setup technically incorrect for the general case.gpt-oss:20b: pass 2026-11-08qwen3.6:27b-mlx: fail (style) 2026-11-07 — [domain objection, downgraded to style] The squeeze theorem requires the inequalities to hold in a neighborhood of the limit point. For x approaching infinity, x is positive, so 1/x > 0. However, the solution fails to explicitly restrict x > 0 or handle the sign of x, which is necessary to correctly bound cos(4x)/x by -1/x and 1/x. Without specifying x > 0, the inequality direction for negative x would be reversed, making the bounds invalid for the general case.gpt-oss:20b: pass 2026-11-07qwen3.6:27b-mlx: fail (style) 2026-11-07 — [domain objection, downgraded to style] The squeeze theorem bounds are incorrect for x -> infinity. Since x is positive, 1/x > 0, so the bounds should be 2 - 1/x and 2 + 1/x. However, the solution claims the expression lies between 2 - 1/x and 2 + 1/x, which is technically correct, but the logic is flawed because it doesn't explicitly handle the sign of x or the inequality direction properly in the context of the limit. More critically, the bounds 2 - 1/x and 2 + 1/x are correct, but the solution fails to explicitly state that x > 0 for large x, which is necessary to ensure 1/x is positive and the inequalities hold as stated. Without this, the 'squeeze' is not rigorously justified. Actually, looking closer, the bounds are correct. The issue is that the solution is too brief and skips the crucial step of dividing by x (which is positive) to establish the inequalities. A student might not realize why the inequalities hold. But is it an error? No, it's just implicit. Let's look for a real error. Ah, the problem is x -> infinity. The bounds are 2 - 1/x and 2 + 1/x. As x -> infinity, 1/x -> 0. The limits are 2. The squeeze theorem applies. Is there a subtle error? The solution says 'the expression lies between 2 - 1/x and 2 + 1/x'. This is true for x > 0. For x -> infinity, x is positive. So the setup is correct. Why would it fail? Maybe the 'unchecked' equations are not the issue. Let's re-read carefully. 'Since -1 <= cos(4x) <= 1, the expression lies between 2 - 1/x and 2 + 1/x.' This step assumes x > 0. If x were negative, the inequalities would flip. Since the limit is x -> infinity, x is positive. This is a standard assumption but should be stated. Is it an error? It's a missing hypothesis. However, usually, this is considered acceptable shorthand. Let's look for a harder error. What if the bounds don't squeeze? They do. What if the limit is wrong? No. Is it misleading? It teaches that you can just pick bounds without checking the sign of the denominator. That is a bad habit. But is it an 'error'? Let's check the severity. If I say 'fail', I need a strong reason. The reason 'missing x>0 condition' is weak for 'error'. Let's look at the bounds again. 2 + cos(4x)/x. Bounds: 2 - 1/|x| <= 2 + cos(4x)/x <= 2 + 1/|x|. For x->infinity, |x|=x. So 2 - 1/x <= ... <= 2 + 1/x. The solution uses 1/x. This is correct for x>0. I will mark it as 'pass' because it is standard calculus shorthand. Wait, the prompt says 'find what is wrong'. If there is nothing wrong, I say pass. Is there anything wrong? The solution is correct. The bounds are correct. The limits are correct. The conclusion is correct. The only 'flaw' is brevity. I will pass it.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.