The squeeze theorem
Problem 1.633 · medium
Use the squeeze theorem to find \( \displaystyle \lim_{x \to 0} x^{3} \sin{\left(\frac{1}{x} \right)} \).
- Since −1 ≤ sin(·) ≤ 1, the expression lies between -x**2*Abs(x) and x**2*Abs(x).
- \[ \lim_{x \to 0}\left(- x^{2} \left|{x}\right|\right) = 0 \]The lower bound's limit.✓ Proved
- \[ \lim_{x \to 0}\left(x^{2} \left|{x}\right|\right) = 0 \]The upper bound's limit, the same.✓ Proved
- Both bounds tend to the same value, so the expression trapped between them does too.
Answer \( 0 \)
Lines: 2 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. 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 (error) — The bounds in step 1 are incorrect; multiplying -1 <= sin(1/x) <= 1 by x^3 yields -|x^3| <= x^3 sin(1/x) <= |x^3|, not bounds involving x^2. Although the limits of the incorrect bounds also happen to be 0, the squeeze theorem requires the function to actually be trapped between the chosen bounds, which is false here.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The bounds in step 1 are incorrect; multiplying -1 <= sin(1/x) <= 1 by x^3 yields -|x^3| <= x^3 sin(1/x) <= |x^3|, not bounds involving x^2. Although the limits of the incorrect bounds also happen to be 0, the squeeze theorem requires the function to actually be trapped between the chosen bounds, which is false here.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer response could not be parsed: { "verdict": "pass", "severity": "", "notes": "The solution correctly identifies the bounds for $x^3 \sin(1/x)$ as $\pm |x|^3$ (written as $\pm x^2 |x|$) and correctly applies the Squeeze Theorem." }gpt-oss:20b: pass 2026-10-14qwen3.6:27b-mlx: fail (style) 2026-10-14 — [domain objection, downgraded to style] The bounds derived in step 1 are incorrect. Since $-1 \le \sin(1/x) \le 1$, multiplying by $x^3$ yields $-|x^3| \le x^3 \sin(1/x) \le |x^3|$. The solution incorrectly uses bounds of order $x^2|x|$ (i.e., $|x|^3$ is correct, but the text says $x^2|Abs(x)|$ which is $|x|^3$, wait. Let's look closer. $x^2 * Abs(x) = |x|^3$. So the bounds are actually correct algebraically. However, the phrasing 'lies between -x**2*Abs(x) and x**2*Abs(x)' is technically correct for the magnitude, but usually squeeze theorem is applied with $-|x^3|$ and $|x^3|$. Let's re-read carefully. $x^3 \sin(1/x)$. Max value is $|x^3|$. The solution says bounds are $\pm x^2 |x| = \pm |x|^3$. This is correct. Why did I think it was wrong? Ah, I misread $x^2$ as the power. $x^2 |x| = |x|^3$. So the bounds are correct. The limits are correct. The conclusion is correct. Is there a subtle error? 'dir=+-' is non-standard notation but the limits are 0. The sentence says 'expression lies between...'. This is true. The logic holds. Wait, is $x^2 |x|$ always equal to $|x^3|$? Yes. Is it equal to $x^3$? No. But the bounds are $\pm |x^3|$. The solution writes $\pm x^2 |x|$. This is a valid representation of $\pm |x^3|$. So the math is correct. Why would it fail? Maybe the notation 'dir=+-' is confusing? Or maybe the prompt implies I should find an error. Let's look at the bounds again. $-1 \le \sin(1/x) \le 1$. Multiply by $x^3$. If $x>0$, $-x^3 \le x^3 \sin(1/x) \le x^3$. If $x<0$, $x^3 \le x^3 \sin(1/x) \le -x^3$ (since multiplying by negative flips inequalities). In both cases, $-|x^3| \le x^3 \sin(1/x) \le |x^3|$. The solution claims the bounds are $-x^2|x|$ and $x^2|x|$. Since $x^2|x| = |x^3|$, the bounds are $-|x^3|$ and $|x^3|$. This is correct. The limits of $\pm |x^3|$ as $x \to 0$ are 0. The squeeze theorem applies. The solution is sound. I will mark it pass.gpt-oss:20b: pass 2026-10-14
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 2026-10-17 with SymPy 1.14.0.