Continuity and types of discontinuity
Problem 1.755 · hard
Find every point where \( \displaystyle f(x) = \frac{3 x^{2} - 3}{x^{2} + 4 x + 3} \) 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.
- \[ x^{2} + 4 x + 3 = \left(x + 1\right) \left(x + 3\right) \]Factor the denominator.✓ Proved
- \[ \frac{3 x^{2} - 3}{x^{2} + 4 x + 3} = \frac{3 x - 3}{x + 3} \]The factor (x + 1) cancels.✓ Proved
- \[ \lim_{x \to -1^+}\left(\frac{3 x^{2} - 3}{x^{2} + 4 x + 3}\right) = -3 \]At x = -1 the limit exists, but f(-1) is undefined: a removable discontinuity.✓ Proved
- \[ \left. 3 x - 3 \right|_{\substack{ x=-3 }} = -12 \]At x = -3 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved
Answer \( \text{removable at } x = -1;\ \text{infinite at } x = -3 \)
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: 'infinite', -1: 'removable'} |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution incorrectly claims the limit at x = -1 is -3; the correct limit is 0. Additionally, the reasoning for the infinite discontinuity at x = -3 is flawed because it evaluates the simplified numerator (which is 0 at x=-3) rather than the original numerator (which is -12), leading to a contradiction in the text.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-25 — The solution incorrectly claims the limit at x = -1 is -3; the correct limit is 0. Additionally, the reasoning for the infinite discontinuity at x = -3 is flawed because it evaluates the simplified numerator (which is 0 at x=-3) rather than the original numerator (which is -12), leading to a contradiction in the text.gpt-oss:20b: pass 2026-10-25qwen3.6:27b-mlx: pass 2026-10-24gpt-oss:20b: pass 2026-10-24qwen3.6:27b-mlx: fail (style) 2026-10-24 — [domain objection, downgraded to style] Step 5 incorrectly substitutes x = -3 into the simplified numerator (3x - 3) to conclude there is an infinite discontinuity. While the conclusion is correct, the reasoning is flawed because one cannot evaluate the simplified expression at a point where the original function is undefined to determine the behavior; one must analyze the limit of the original function or the simplified form's limit behavior (which goes to infinity), not just substitute into the numerator.gpt-oss:20b: pass 2026-10-24
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.