Continuity and types of discontinuity
Problem 1.760 · hard
Find every point where \( \displaystyle f(x) = \frac{2 x + 6}{x^{2} + 2 x - 3} \) 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.
- \[ x^{2} + 2 x - 3 = \left(x - 1\right) \left(x + 3\right) \]Factor the denominator.✓ Proved
- \[ \frac{2 x + 6}{x^{2} + 2 x - 3} = \frac{2}{x - 1} \]The factor (x + 3) cancels.✓ Proved
- \[ \lim_{x \to -3^+}\left(\frac{2 x + 6}{x^{2} + 2 x - 3}\right) = - \frac{1}{2} \]At x = -3 the limit exists, but f(-3) is undefined: a removable discontinuity.✓ Proved
- \[ \left. 2 \right|_{\substack{ x=1 }} = 2 \]At x = 1 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved
Answer \( \text{removable at } x = -3;\ \text{infinite at } x = 1 \)
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. 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 | ✓ 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', 1: 'infinite'} |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — Step 5 incorrectly substitutes x=1 into the simplified numerator '2' instead of the original numerator '2x+6' (which is 8). While the conclusion of an infinite discontinuity is correct, the reasoning 'numerator is not 0' relies on a calculation error that would fail for other functions.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-25 — Step 5 incorrectly substitutes x=1 into the simplified numerator '2' instead of the original numerator '2x+6' (which is 8). While the conclusion of an infinite discontinuity is correct, the reasoning 'numerator is not 0' relies on a calculation error that would fail for other functions.gpt-oss:20b: pass 2026-10-25qwen3.6:27b-mlx: fail (style) 2026-10-24 — [domain objection, downgraded to style] Step 5 incorrectly substitutes x=1 into the simplified expression 2/(x-1) to get 2, which is undefined. The logic that the numerator is non-zero at x=1 is correct, but the calculation shown is mathematically invalid and misleading.gpt-oss:20b: fail (error) 2026-10-24 — Step 5 incorrectly evaluates the numerator at x=1 (it should be 8, not 2) and does not justify that the limit is infinite; the conclusion about an infinite discontinuity is therefore unsupported.qwen3.6:27b-mlx: pass 2026-10-24gpt-oss:20b: fail (error) 2026-10-24 — Step 5 incorrectly evaluates the numerator at x=1; it should be 2(1)+6=8, not 2. The conclusion about an infinite discontinuity is correct, but the calculation is wrong.
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 2026-10-25 with SymPy 1.14.0.