Home›Calculus 1›Continuity and types of discontinuity›Problem 1.398

Continuity and types of discontinuity

Problem 1.398 · hard

Find every point where \( \displaystyle f(x) = \frac{8 x + 8}{x^{2} - 4 x - 5} \) 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
\[ x^{2} - 4 x - 5 = \left(x - 5\right) \left(x + 1\right) \]
Factor the denominator.✓ Proved
\[ \frac{8 x + 8}{x^{2} - 4 x - 5} = \frac{8}{x - 5} \]
The factor (x + 1) cancels.✓ Proved
\[ \lim_{x \to -1^+}\left(\frac{8 x + 8}{x^{2} - 4 x - 5}\right) = - \frac{4}{3} \]
At x = -1 the limit exists, but f(-1) is undefined: a removable discontinuity.✓ Proved
\[ \left. 8 \right|_{\substack{ x=5 }} = 8 \]
At x = 5 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved

Answer \( \text{removable at } x = -1;\ \text{infinite at } x = 5 \)

✓ 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: This is a sentence stating 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 not an algebraic equivalence.
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: {-1: 'removable', 5: 'infinite'}

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "Step 4 claims the limit at x = -1 is -4/3, but substituting x = -1 into the simplified expression 8/(x-5) yields 8/(-6) = -4/3. Wait, 8/(-6) is indee

Senior review claude-sonnet-5-5, 2026-10-07: pass — Factorization, cancellation, the limit -4/3 at x=-1, and the infinite discontinuity at x=5 are all correct and match the stated answer. Line 5's Subs(8, x, 5) is terse (it evaluates the cancelled numerator 8), but it is true and supports the right conclusion.

qwen3.6:27b-mlx: dismiss — The objection itself re-derives that every line is correct (factorization, limit -4/3, classification). The only complaint is the odd Subs(8, x, 5) = 8 notation, which is a true statement about the cancelled numerator and not a false claim, a wrong rule name, or a double step.

Every verdict on record (5)

gpt-oss:20b: pass 2026-10-07
qwen3.6:27b-mlx: inconclusive 2026-10-07 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "Step 4 claims the limit at x = -1 is -4/3, but substituting x = -1 into the simplified expression 8/(x-5) yields 8/(-6) = -4/3. Wait, 8/(-6) is indee
claude-sonnet-5-5: pass 2026-10-07 — Factorization, cancellation, the limit -4/3 at x=-1, and the infinite discontinuity at x=5 are all correct and match the stated answer. Line 5's Subs(8, x, 5) is terse (it evaluates the cancelled numerator 8), but it is true and supports the right conclusion.
qwen3.6:27b-mlx: fail (error) 2026-10-07 — The limit at x = -1 is calculated incorrectly; the simplified function 8/(x-5) evaluates to -4/3, but the original function simplifies to 8/(x-5) only after canceling (x+1), so the limit is indeed -4/3. Wait, let me re-evaluate. f(x) = 8(x+1)/((x-5)(x+1)). Limit as x->-1 is 8(-1+1)/((-1-5)(-1+1)) -> 0/0. Cancel (x+1): 8/(x-5). Limit as x->-1 is 8/(-1-5) = 8/-6 = -4/3. The calculation in line 4 is correct. However, line 5 says 'Subs(8, x, 5) = 8'. This is syntactically weird and conceptually wrong. It substitutes x=5 into the numerator 8x+8? No, 8(5)+8 = 48. It substitutes into the simplified numerator 8? If it means the numerator of the simplified form is 8, then 8 != 0, so infinite discontinuity. But the notation 'Subs(8, x, 5)' is nonsensical because 8 does not depend on x. More importantly, the classification is correct, but the reasoning in line 5 is poorly expressed. Let's look closer. Is there an error? Line 4 says limit is -4/3. Line 5 says 'Subs(8, x, 5) = 8'. This likely refers to the numerator of the simplified fraction 8/(x-5). The numerator is 8. At x=5, the denominator is 0. Since the numerator is non-zero (8), it is an infinite discontinuity. This is correct logic. Is there a mistake? Let's check the limit calculation again. 8/(-6) = -4/3. Correct. Let's check the infinite discontinuity. Denominator zero at 5. Numerator 8(5)+8 = 48 != 0. Correct. Why would this fail? Maybe the prompt implies I should find a subtle error. Let's look at line 3. 'The factor (x+1) cancels'. This is valid for x != -1. Line 4 uses this to find the limit. This is standard. Line 5 uses the simplified form to check the numerator at x=5. This is also standard. Is there a style issue? 'Subs(8, x, 5)' is very poor notation. It suggests substituting x into the constant 8. A better sentence would be 'The numerator of the simplified function is 8, which is non-zero at x=5'. This is misleading/poor style. But is it an error? It leads to the correct conclusion. Let's look for a harder error. What if the limit at -1 was calculated wrong? 8/(-6) = -4/3. Correct. What if the classification is wrong? Removable at -1, Infinite at 5. Correct. Is there a missing discontinuity? No. Is the factorization correct? (x-5)(x+1) = x^2 -4x -5. Correct. Is the numerator factorization correct? 8x+8 = 8(x+1). Correct. So the math is correct. The only issue is the notation in line 5. 'Subs(8, x, 5)' is not a standard mathematical statement. It's a computer algebra command that doesn't make sense in this context (substituting into a constant). This is misleading to a student who might think you substitute into the original numerator 8x+8 and get 8? No, 8(5)+8=48. If the student thinks the numerator is 8, they might think the original numerator was 8. This is misleading. Severity: misleading.
gpt-oss:20b: pass 2026-10-07

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 2026-10-07 with SymPy 1.14.0.