Continuity and types of discontinuity
Problem 1.542 · hard
Find every point where \( \displaystyle f(x) = \frac{2 x - 10}{x^{2} - x - 20} \) 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.Reviewed
- \[ x^{2} - x - 20 = \left(x - 5\right) \left(x + 4\right) \]Factor the denominator.✓ Proved
- \[ \frac{2 x - 10}{x^{2} - x - 20} = \frac{2}{x + 4} \]The factor (x − 5) cancels.✓ Proved
- \[ \lim_{x \to 5^+}\left(\frac{2 x - 10}{x^{2} - x - 20}\right) = \frac{2}{9} \]At x = 5 the limit exists, but f(5) is undefined: a removable discontinuity.✓ Proved
- \[ \left. 2 \right|_{\substack{ x=-4 }} = 2 \]At x = -4 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved
Answer \( \text{removable at } x = 5;\ \text{infinite at } x = -4 \)
✓ 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: Line 1 is a prose statement of a standard fact: a rational function is continuous wherever its denominator is nonzero, so only the denominator's zeros can be discontinuities. It is true and is not an algebraic equivalence a script could check. |
| 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: 'infinite', 5: 'removable'} |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — Line 5 incorrectly substitutes x = -4 into the simplified numerator 2, yielding 2, instead of evaluating the original numerator 2x - 10 at x = -4, which yields -18. While the classification of the discontinuity as infinite is correct, the reasoning provided in the sentence is mathematically false.
Senior review claude-sonnet-5-5, 2026-10-11: pass — The factorization, cancellation, removable limit 2/9 at x=5, and infinite discontinuity at x=-4 are all correct and match the stated answer. Line 5's 'Subs(2, x, -4) = 2' is an awkward way to show the reduced numerator is nonzero, but it is true and not misleading.
gpt-oss:20b: dismiss — Line 5 substitutes into the numerator 2 of the reduced form 2/(x+4), which is a true evaluation. The claim that the numerator is nonzero holds for the original numerator (-18) as well, so nothing false is stated.qwen3.6:27b-mlx: dismiss — The line evaluates only the constant numerator 2 at x=-4, not the whole expression 2/(x+4). The conclusion that the denominator vanishes while the numerator is nonzero is correct and supports an infinite discontinuity.
Every verdict on record (5)
qwen3.6:27b-mlx: fail (error) 2026-10-11 — Line 5 incorrectly substitutes x = -4 into the simplified numerator 2, yielding 2, instead of evaluating the original numerator 2x - 10 at x = -4, which yields -18. While the classification of the discontinuity as infinite is correct, the reasoning provided in the sentence is mathematically false.gpt-oss:20b: pass 2026-10-11claude-sonnet-5-5: pass 2026-10-11 — The factorization, cancellation, removable limit 2/9 at x=5, and infinite discontinuity at x=-4 are all correct and match the stated answer. Line 5's 'Subs(2, x, -4) = 2' is an awkward way to show the reduced numerator is nonzero, but it is true and not misleading.gpt-oss:20b: fail (error) 2026-10-11 — Step 5 incorrectly substitutes the numerator at x=-4; the numerator is -18, not 2, but the conclusion about an infinite discontinuity is still correct. The statement should read "At x=-4 the numerator equals -18, while the denominator vanishes, so |f(x)|→∞, giving an infinite discontinuity."qwen3.6:27b-mlx: fail (style) 2026-10-11 — [domain objection, downgraded to style] Step 5 incorrectly substitutes x = -4 into the simplified expression 2/(x+4) to get 2, which is undefined. This contradicts the conclusion of an infinite discontinuity and teaches incorrect evaluation of limits at vertical asymptotes.
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.