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Continuity and types of discontinuity

Problem 1.283 · hard

Find every point where \( \displaystyle f(x) = \frac{x - 1}{x^{2} + 2 x - 3} \) 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.
Reviewed
\[ x^{2} + 2 x - 3 = \left(x - 1\right) \left(x + 3\right) \]
Factor the denominator.✓ Proved
\[ \frac{x - 1}{x^{2} + 2 x - 3} = \frac{1}{x + 3} \]
The factor (x − 1) cancels.✓ Proved
\[ \lim_{x \to 1^+}\left(\frac{x - 1}{x^{2} + 2 x - 3}\right) = \frac{1}{4} \]
At x = 1 the limit exists, but f(1) is undefined: a removable discontinuity.✓ Proved
\[ \left. 1 \right|_{\substack{ x=-3 }} = 1 \]
At x = -3 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved

Answer \( \text{removable at } x = 1;\ \text{infinite at } x = -3 \)

✓ 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 sentence, not an equation: a rational function is continuous wherever its denominator is nonzero, so only the denominator's zeros can be discontinuities. This is a standard true fact.
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: pass
qwen3.6:27b-mlx: fail (error) — Step 5 contains a fatal algebraic error: substituting x=1 into the expression (x-1)/(x+3) yields 0, not 1. The correct numerator at x=-3 is -4, leading to an infinite discontinuity, but the stated calculation is wrong.

Senior review claude-sonnet-5-5, 2026-10-05: pass — Factoring, cancellation, the removable limit 1/4 at x = 1 and the infinite discontinuity at x = -3 are all correct and match the stated answer. Line 5's Subs(1, x, -3) = 1 is an odd way to show the numerator is nonzero, but it is true and not misleading.

gpt-oss:20b: dismiss — Line 5 substitutes x = -3 into the constant numerator 1 of the reduced form, not into f, and the true value 1 shows a nonzero numerator over a zero denominator, which supports the infinite discontinuity.
qwen3.6:27b-mlx: dismiss — Line 5 does not substitute x = 1 into 1/(x+3); Subs(1, x, -3) = 1 is a true statement, and the narrative claim that the numerator is nonzero at x = -3 holds for the original numerator (-4) and the reduced one (1).

Every verdict on record (5)

qwen3.6:27b-mlx: fail (error) 2026-10-05 — Step 5 contains a fatal algebraic error: substituting x=1 into the expression (x-1)/(x+3) yields 0, not 1. The correct numerator at x=-3 is -4, leading to an infinite discontinuity, but the stated calculation is wrong.
gpt-oss:20b: pass 2026-10-05
claude-sonnet-5-5: pass 2026-10-05 — Factoring, cancellation, the removable limit 1/4 at x = 1 and the infinite discontinuity at x = -3 are all correct and match the stated answer. Line 5's Subs(1, x, -3) = 1 is an odd way to show the numerator is nonzero, but it is true and not misleading.
qwen3.6:27b-mlx: fail (error) 2026-10-05 — Line 5 contains a severe algebraic error: substituting x=1 into the expression 1/(x+3) yields 1/4, not 1. Furthermore, the reasoning for the infinite discontinuity at x=-3 is flawed because it references the simplified numerator (which is 1) rather than the original numerator (which is -4), although the conclusion about the infinite discontinuity is correct, the justification is mathematically incorrect.
gpt-oss:20b: fail (style) 2026-10-05 — [domain objection, downgraded to style] Step 5 incorrectly substitutes x = -3 into the function; the expression is undefined there, and the correct conclusion is that the limit does not exist (infinite discontinuity).

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 2026-10-05 with SymPy 1.14.0.