Continuity and types of discontinuity
Problem 1.752 · hard
Find every point where \( \displaystyle f(x) = \frac{40 - 8 x}{x^{2} - 3 x - 10} \) 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.
- \[ x^{2} - 3 x - 10 = \left(x - 5\right) \left(x + 2\right) \]Factor the denominator.✓ Proved
- \[ \frac{40 - 8 x}{x^{2} - 3 x - 10} = - \frac{8}{x + 2} \]The factor (x − 5) cancels.✓ Proved
- \[ \lim_{x \to 5^+}\left(\frac{40 - 8 x}{x^{2} - 3 x - 10}\right) = - \frac{8}{7} \]At x = 5 the limit exists, but f(5) 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 \( \text{removable at } x = 5;\ \text{infinite at } x = -2 \)
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. 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: {-2: 'infinite', 5: 'removable'} |
Reviewers
gpt-oss:20b: fail (style) — [domain objection, downgraded to style] The solution incorrectly evaluates the function at x = -2 by writing Subs(-8, x, -2) = -8, which is nonsensical. It also fails to explain that the function is undefined at x = -2 and that the limit does not exist, leading to an infinite discontinuity.qwen3.6:27b-mlx: fail (error) — Step 5 incorrectly substitutes x = -2 into the simplified numerator -8 to conclude it is non-zero, ignoring that the original numerator 40 - 8x is actually 0 at x = -2. This leads to the false classification of an infinite discontinuity instead of a removable one.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-25 — Step 5 incorrectly substitutes x = -2 into the simplified numerator -8 to conclude it is non-zero, ignoring that the original numerator 40 - 8x is actually 0 at x = -2. This leads to the false classification of an infinite discontinuity instead of a removable one.gpt-oss:20b: fail (style) 2026-10-25 — [domain objection, downgraded to style] The solution incorrectly evaluates the function at x = -2 by writing Subs(-8, x, -2) = -8, which is nonsensical. It also fails to explain that the function is undefined at x = -2 and that the limit does not exist, leading to an infinite discontinuity.qwen3.6:27b-mlx: fail (error) 2026-10-24 — Step 5 incorrectly substitutes x = -2 into the simplified numerator -8 to claim it is non-zero, ignoring that the original numerator 40 - 8x is also 0 at x = -2. Since both numerator and denominator are zero at x = -2, the limit exists (finite), making it a removable discontinuity, not infinite.gpt-oss:20b: fail (misleading) 2026-10-24 — The solution incorrectly evaluates the numerator at x = -2, writing Subs(-8, x, -2) = -8. The numerator is 40–8(–2)=56, not –8. Although the conclusion that the discontinuity at x = –2 is infinite is correct, the erroneous evaluation could mislead a student about how to compute limits at a vertical asymptote.qwen3.6:27b-mlx: fail (error) 2026-10-24 — Line 5 incorrectly substitutes x = -2 into the simplified numerator -8, claiming the result is -8, to argue the numerator is non-zero. While the conclusion (infinite discontinuity) is correct, the reasoning is flawed because the original numerator at x=-2 is 24, not -8. One must check the original numerator or note that the cancelled factor (x-5) is non-zero at x=-2, but simply stating 'Subs(-8, x, -2) = -8' as the justification for the numerator's value at the discontinuity is mathematically imprecise and misleading regarding the function's behavior.gpt-oss:20b: fail (error) 2026-10-24 — Step 5 incorrectly evaluates the function at x = -2; it should discuss the limit of -8/(x+2) as x→-2, which diverges, rather than substituting -8 for the numerator.
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.