Home›Calculus 1›Continuity and types of discontinuity›Problem 1.281

Continuity and types of discontinuity

Problem 1.281 · hard

Find every point where \( \displaystyle f(x) = \frac{2 x^{2} + 4 x - 16}{x^{2} + x - 12} \) 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
\[ x^{2} + x - 12 = \left(x - 3\right) \left(x + 4\right) \]
Factor the denominator.✓ Proved
\[ \frac{2 x^{2} + 4 x - 16}{x^{2} + x - 12} = \frac{2 x - 4}{x - 3} \]
The factor (x + 4) cancels.✓ Proved
\[ \lim_{x \to -4^+}\left(\frac{2 x^{2} + 4 x - 16}{x^{2} + x - 12}\right) = \frac{12}{7} \]
At x = -4 the limit exists, but f(-4) is undefined: a removable discontinuity.✓ Proved
\[ \left. 2 x - 4 \right|_{\substack{ x=3 }} = 2 \]
At x = 3 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved

Answer \( \text{removable at } x = -4;\ \text{infinite at } x = 3 \)

✓ 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: A prose statement that rational functions are continuous wherever the denominator is nonzero, so only denominator zeros can be discontinuities. This is a true standard fact, with no algebraic equivalence to 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: 'removable', 3: 'infinite'}

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass

Senior review claude-sonnet-5-5, 2026-10-05: pass — The factorization, cancellation, limit 12/7 at x=-4 (removable), and nonzero numerator at x=3 with zero denominator (infinite) are all correct and match the stated answer. Line 1 is a true sentence about rational functions.

qwen3.6:27b-mlx: dismiss — Since 2x^2+4x-16 = 2(x+4)(x-2), f = (2x-4)/(x-3), and at x=-4 this gives (-12)/(-7) = 12/7, so the stated limit is correct and the claimed -12/7 is a sign error. Step 5 only evaluates the reduced numerator at 3 (giving 2, nonzero) to justify unbounded growth, and never claims the limit is 2.

Every verdict on record (5)

qwen3.6:27b-mlx: pass 2026-10-05
gpt-oss:20b: pass 2026-10-05
claude-sonnet-5-5: pass 2026-10-05 — The factorization, cancellation, limit 12/7 at x=-4 (removable), and nonzero numerator at x=3 with zero denominator (infinite) are all correct and match the stated answer. Line 1 is a true sentence about rational functions.
qwen3.6:27b-mlx: fail (error) 2026-10-05 — The limit calculation in step 4 is incorrect; the limit as x approaches -4 is 12/7, but the simplified expression (2x-4)/(x-3) evaluated at -4 yields -12/7. Additionally, step 5 incorrectly substitutes into the simplified numerator only, ignoring the denominator, to claim the limit is 2, whereas the limit is actually infinite.
gpt-oss:20b: pass 2026-10-05

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.