Continuity and types of discontinuity
Problem 1.396 · hard
Find every point where \( \displaystyle f(x) = \frac{- 2 x^{2} + 4 x + 30}{x^{2} - 3 x - 10} \) is discontinuous, and classify each discontinuity as removable, a jump, or infinite.
- A rational function is continuous wherever its denominator is not zero, so only the zeros of the denominator can be discontinuities.
- \[ x^{2} - 3 x - 10 = \left(x - 5\right) \left(x + 2\right) \]Factor the denominator.✓ Proved
- \[ \frac{- 2 x^{2} + 4 x + 30}{x^{2} - 3 x - 10} = \frac{- 2 x - 6}{x + 2} \]The factor (x − 5) cancels.✓ Proved
- \[ \lim_{x \to 5^+}\left(\frac{- 2 x^{2} + 4 x + 30}{x^{2} - 3 x - 10}\right) = - \frac{16}{7} \]At x = 5 the limit exists, but f(5) is undefined: a removable discontinuity.✓ Proved
- \[ \left. - 2 x - 6 \right|_{\substack{ x=-2 }} = -2 \]At x = -2 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved
Answer \( \text{removable at } x = 5;\ \text{infinite at } x = -2 \)
Lines: 4 proved, 1 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.
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 | ✓ 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: {-2: 'infinite', 5: 'removable'} |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The limit calculation at x=5 is incorrect; the limit is -16/3, not -16/7. Additionally, the reasoning for the infinite discontinuity at x=-2 is flawed because it relies on the simplified expression's numerator (-2) rather than the original numerator (24), although the conclusion is correct.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: fail (error) 2026-10-07 — The limit calculation at x=5 is incorrect; the limit is -16/3, not -16/7. Additionally, the reasoning for the infinite discontinuity at x=-2 is flawed because it relies on the simplified expression's numerator (-2) rather than the original numerator (24), although the conclusion is correct.qwen3.6:27b-mlx: inconclusive 2026-10-07 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "Step 4 claims the limit at x=5 is -16/7, but substituting x=5 into the simplified expression (-2x-6)/(x+2) yields -16/7 only if the denominator is 7gpt-oss:20b: fail (error) 2026-10-07 — The solution incorrectly cancels the factor (x‑5) from the numerator, yielding the wrong simplified form — the numerator should be “-2(x+3)”, not “-2x-6”. This mis‑cancellation undermines the correctness of the subsequent analysis.
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 2026-10-07 with SymPy 1.14.0.