Continuity and types of discontinuity
Problem 1.539 · hard
Find every point where \( \displaystyle f(x) = \frac{8 x + 24}{x^{2} + x - 6} \) 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.
- \[ x^{2} + x - 6 = \left(x - 2\right) \left(x + 3\right) \]Factor the denominator.✓ Proved
- \[ \frac{8 x + 24}{x^{2} + x - 6} = \frac{8}{x - 2} \]The factor (x + 3) cancels.✓ Proved
- \[ \lim_{x \to -3^+}\left(\frac{8 x + 24}{x^{2} + x - 6}\right) = - \frac{8}{5} \]At x = -3 the limit exists, but f(-3) is undefined: a removable discontinuity.✓ Proved
- \[ \left. 8 \right|_{\substack{ x=2 }} = 8 \]At x = 2 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved
Answer \( \text{removable at } x = -3;\ \text{infinite at } x = 2 \)
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: {-3: 'removable', 2: 'infinite'} |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — Line 5 incorrectly substitutes x=2 into the simplified numerator 8 (which is 8) and claims this proves an infinite discontinuity. The correct reasoning is that the limit of the simplified function 8/(x-2) as x->2 is infinite because the denominator approaches 0 while the numerator approaches 8. The sentence 'Subs(8, x, 2) = 8' is algebraically trivial and does not constitute a proof of the limit behavior; it misrepresents the logic for classifying the discontinuity.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-11 — Line 5 incorrectly substitutes x=2 into the simplified numerator 8 (which is 8) and claims this proves an infinite discontinuity. The correct reasoning is that the limit of the simplified function 8/(x-2) as x->2 is infinite because the denominator approaches 0 while the numerator approaches 8. The sentence 'Subs(8, x, 2) = 8' is algebraically trivial and does not constitute a proof of the limit behavior; it misrepresents the logic for classifying the discontinuity.gpt-oss:20b: pass 2026-10-11gpt-oss:20b: pass 2026-10-11qwen3.6:27b-mlx: pass 2026-10-11
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 2026-10-11 with SymPy 1.14.0.