Continuity and types of discontinuity
Problem 1.197 · hard
- A rational function is continuous wherever its denominator is not zero, so only the zeros of the denominator can be discontinuities.Reviewed
- \[ x^{2} + x - 12 = \left(x - 3\right) \left(x + 4\right) \]Factor the denominator.✓ Proved
- \[ \frac{3 x^{2} + 9 x - 12}{x^{2} + x - 12} = \frac{3 x - 3}{x - 3} \]The factor (x + 4) cancels.✓ Proved
- \[ \lim_{x \to -4^+}\left(\frac{3 x^{2} + 9 x - 12}{x^{2} + x - 12}\right) = \frac{15}{7} \]At x = -4 the limit exists, but f(-4) is undefined: a removable discontinuity.✓ Proved
- \[ \left. 3 x - 3 \right|_{\substack{ x=3 }} = 6 \]At x = 3 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, not an equation. It correctly states that a rational function is continuous wherever its denominator is nonzero, so only the zeros of the denominator can be discontinuities. |
| 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: {-4: 'removable', 3: 'infinite'} |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution incorrectly identifies the removable discontinuity at x = -4. The factor (x+4) cancels, leaving a denominator of (x-3), which is non-zero at x = -4, meaning the function is actually continuous at x = -4. The infinite discontinuity is correctly identified at x = 3, but the classification for x = -4 is wrong.
Senior review claude-sonnet-5-5, 2026-10-04: pass — The math is correct. The denominator factors as (x-3)(x+4), the numerator as 3(x+4)(x-1), so -4 is removable with limit 15/7. At x=3 the numerator is 6, not 0, so the discontinuity is infinite. The stated answer matches.
qwen3.6:27b-mlx: dismiss — The objection is false: 3x^2+9x-12 = 3(x+4)(x-1), so (x+4) does cancel, the limit at -4 is 15/7, and x=-4 is removable.
Every verdict on record (5)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution incorrectly identifies the removable discontinuity at x = -4. The factor (x+4) cancels, leaving a denominator of (x-3), which is non-zero at x = -4, meaning the function is actually continuous at x = -4. The infinite discontinuity is correctly identified at x = 3, but the classification for x = -4 is wrong.gpt-oss:20b: pass 2026-10-04claude-sonnet-5-5: pass 2026-10-04 — The math is correct. The denominator factors as (x-3)(x+4), the numerator as 3(x+4)(x-1), so -4 is removable with limit 15/7. At x=3 the numerator is 6, not 0, so the discontinuity is infinite. The stated answer matches.qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution incorrectly identifies x = -4 as a removable discontinuity. The factor (x+4) is in the denominator but does not cancel with the numerator (which factors to 3(x+4)(x-1)), meaning x=-4 is actually an infinite discontinuity. The simplification in step 3 is algebraically incorrect.gpt-oss:20b: pass 2026-10-04
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 2026-10-04 with SymPy 1.14.0.