Continuity and types of discontinuity
Problem 1.484 · hard
- A rational function is continuous wherever its denominator is not zero, so only the zeros of the denominator can be discontinuities.Reviewed
- \[ x^{2} - 3 x + 2 = \left(x - 2\right) \left(x - 1\right) \]Factor the denominator.✓ Proved
- \[ \frac{30 - 15 x}{x^{2} - 3 x + 2} = - \frac{15}{x - 1} \]The factor (x − 2) cancels.✓ Proved
- \[ \lim_{x \to 2^+}\left(\frac{30 - 15 x}{x^{2} - 3 x + 2}\right) = -15 \]At x = 2 the limit exists, but f(2) is undefined: a removable discontinuity.✓ Proved
- \[ \left. -15 \right|_{\substack{ x=1 }} = -15 \]At x = 1 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Reviewed | claude-sonnet-5-5 | claude-sonnet-5-5: This is a sentence stating the standard fact that a rational function is continuous wherever its denominator is nonzero. It is true, and it sets up the analysis of the denominator's zeros. |
| 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | zeros of the denominator found by solve, each probed at ±1e-9: {1: 'infinite', 2: 'removable'} |
Reviewers
gpt-oss:20b: fail (error) — Sentence 5 incorrectly evaluates the function at x=1. The limit of f(x) as x→1 is infinite, not –15. The statement ‘Subs(-15, x, 1) = -15’ is false and misleads the reader about the nature of the discontinuity.qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] Step 5 incorrectly substitutes x=1 into the simplified expression -15/(x-1) to get -15, which is undefined, not -15. The reasoning that |f(x)| grows without bound is correct for an infinite discontinuity, but the equation shown is mathematically false and misleading.
Senior review claude-sonnet-5-5, 2026-10-09: pass — The factoring, the cancellation, and the classification are all correct: removable at x=2 (limit -15) and infinite at x=1 (reduced numerator -15 is nonzero over a vanishing denominator). Line 5's equation is a clumsy way to show the numerator is nonzero, but it is true and its explanation is sound.
gpt-oss:20b: dismiss — The line substitutes x=1 into the constant -15 (the reduced numerator), which is a true, if trivial, equation. It does not substitute -15 for x, and the accompanying nonzero-numerator reasoning, which the objection itself accepts, is correct.
Every verdict on record (5)
gpt-oss:20b: fail (error) 2026-10-09 — Sentence 5 incorrectly evaluates the function at x=1. The limit of f(x) as x→1 is infinite, not –15. The statement ‘Subs(-15, x, 1) = -15’ is false and misleads the reader about the nature of the discontinuity.qwen3.6:27b-mlx: fail (style) 2026-10-09 — [domain objection, downgraded to style] Step 5 incorrectly substitutes x=1 into the simplified expression -15/(x-1) to get -15, which is undefined, not -15. The reasoning that |f(x)| grows without bound is correct for an infinite discontinuity, but the equation shown is mathematically false and misleading.claude-sonnet-5-5: pass 2026-10-09 — The factoring, the cancellation, and the classification are all correct: removable at x=2 (limit -15) and infinite at x=1 (reduced numerator -15 is nonzero over a vanishing denominator). Line 5's equation is a clumsy way to show the numerator is nonzero, but it is true and its explanation is sound.gpt-oss:20b: fail (error) 2026-10-09 — Line 5 contains an incorrect equation: substituting –15 for x is nonsensical and does not justify the infinite discontinuity at x=1. The reasoning about the numerator being non‑zero at x=1 is correct, but the equation itself is erroneous.qwen3.6:27b-mlx: pass 2026-10-09
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/continuity_classify, checked 2026-10-09 with SymPy 1.14.0.