∫Calc Practice
Home›Calculus 1›Continuity and types of discontinuity›Problem 1.483

Continuity and types of discontinuity

Problem 1.483 · hard

Find every point where \( \displaystyle f(x) = \frac{- 2 x - 2}{x^{2} - 2 x - 3} \) is discontinuous, and classify each discontinuity as removable, a jump, or infinite.
  1. A rational function is continuous wherever its denominator is not zero, so only the zeros of the denominator can be discontinuities.
    Reviewed
  2. \[ x^{2} - 2 x - 3 = \left(x - 3\right) \left(x + 1\right) \]
    Factor the denominator.✓ Proved
  3. \[ \frac{- 2 x - 2}{x^{2} - 2 x - 3} = - \frac{2}{x - 3} \]
    The factor (x + 1) cancels.✓ Proved
  4. \[ \lim_{x \to -1^+}\left(\frac{- 2 x - 2}{x^{2} - 2 x - 3}\right) = \frac{1}{2} \]
    At x = -1 the limit exists, but f(-1) is undefined: a removable discontinuity.✓ Proved
  5. \[ \left. -2 \right|_{\substack{ x=3 }} = -2 \]
    At x = 3 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved
Answer \( \text{removable at } x = -1;\ \text{infinite at } x = 3 \)

✓ 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
LineStatusChecked byDetail
1Reviewedclaude-sonnet-5-5claude-sonnet-5-5: This is a sentence, not an equation. It states a standard fact: a rational function is continuous wherever its denominator is nonzero, so only the zeros of x^2-2x-3 can be discontinuities.
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0zeros of the denominator found by solve, each probed at ±1e-9: {-1: 'removable', 3: 'infinite'}

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (error) — Step 5 incorrectly evaluates the numerator of the simplified function at x=3 as -2, whereas the numerator is the constant -2, so the limit is indeed infinite, but the reasoning 'Subs(-2, x, 3) = -2' is confusingly phrased and technically refers to the simplified numerator. More critically, the classification of the discontinuity at x=3 relies on the limit of the simplified function -2/(x-3). As x->3, the denominator goes to 0, so the limit is infinite. The step says 'numerator is not 0', which is true for the simplified form, but the original numerator at x=3 is -8. The logic holds, but the phrasing in step 5 is sloppy. However, the main error is in Step 3: it states the factor cancels, implying the function is equal to -2/(x-3) everywhere, which is false at x=-1. While standard shorthand, it's imprecise. But the biggest issue is Step 5: 'Subs(-2, x, 3) = -2' is not a standard notation for checking the numerator of the simplified fraction. It should check the limit behavior. Actually, looking closer, Step 5 claims 'At x=3 the numerator is not 0'. In the original function, numerator is -8. In the simplified, it's -2. Both are non-zero. The conclusion is correct. Is there an error? Step 4 calculates the limit from the right. For a removable discontinuity, the two-sided limit must exist. Since the simplified function is continuous at -1, the two-sided limit exists. The one-sided limit check is sufficient but incomplete notation. However, the verdict should be pass if the math is correct. Let's look for a real error. Step 3: 'The factor (x+1) cancels'. This is an algebraic simplification valid for x != -1. Step 4 uses this to find the limit. Step 5 uses the simplified form to find the behavior at x=3. The logic is sound. Why would it fail? Maybe the classification? Removable at -1, Infinite at 3. Correct. Is there a subtle error? 'Subs(-2, x, 3) = -2'. This line is marked 'proved'. It's just substituting 3 into -2. It's trivial. The sentence says 'numerator is not 0'. This is true. The conclusion 'infinite discontinuity' is correct. I see no mathematical error. The solution is correct. Wait, let me re-read the prompt instructions. 'Find every point... classify'. The solution finds -1 and 3. Classifies them correctly. The steps are logically sound. I will mark it as pass.

Senior review claude-sonnet-5-5, 2026-10-09: pass — The factorization, the cancellation, the removable discontinuity at x=-1 (limit 1/2) and the infinite discontinuity at x=3 are all correct and match the stated answer. Line 5's Subs(-2, x, 3) = -2 is a trivial display, and the wording could say that the reduced numerator is nonzero, but it is not false.

  • qwen3.6:27b-mlx: dismiss — Line 5 evaluates the numerator -2 of the reduced form -2/(x-3) at x=3, which is a true and nonzero value. Line 3 already justified the reduction, so the arithmetic is correct, and the classification as an infinite discontinuity is right (the original numerator is -8 there, also nonzero).
Every verdict on record (5)
  • gpt-oss:20b: pass 2026-10-09
  • qwen3.6:27b-mlx: fail (error) 2026-10-09 — Step 5 incorrectly evaluates the numerator of the simplified function at x=3 as -2, whereas the numerator is the constant -2, so the limit is indeed infinite, but the reasoning 'Subs(-2, x, 3) = -2' is confusingly phrased and technically refers to the simplified numerator. More critically, the classification of the discontinuity at x=3 relies on the limit of the simplified function -2/(x-3). As x->3, the denominator goes to 0, so the limit is infinite. The step says 'numerator is not 0', which is true for the simplified form, but the original numerator at x=3 is -8. The logic holds, but the phrasing in step 5 is sloppy. However, the main error is in Step 3: it states the factor cancels, implying the function is equal to -2/(x-3) everywhere, which is false at x=-1. While standard shorthand, it's imprecise. But the biggest issue is Step 5: 'Subs(-2, x, 3) = -2' is not a standard notation for checking the numerator of the simplified fraction. It should check the limit behavior. Actually, looking closer, Step 5 claims 'At x=3 the numerator is not 0'. In the original function, numerator is -8. In the simplified, it's -2. Both are non-zero. The conclusion is correct. Is there an error? Step 4 calculates the limit from the right. For a removable discontinuity, the two-sided limit must exist. Since the simplified function is continuous at -1, the two-sided limit exists. The one-sided limit check is sufficient but incomplete notation. However, the verdict should be pass if the math is correct. Let's look for a real error. Step 3: 'The factor (x+1) cancels'. This is an algebraic simplification valid for x != -1. Step 4 uses this to find the limit. Step 5 uses the simplified form to find the behavior at x=3. The logic is sound. Why would it fail? Maybe the classification? Removable at -1, Infinite at 3. Correct. Is there a subtle error? 'Subs(-2, x, 3) = -2'. This line is marked 'proved'. It's just substituting 3 into -2. It's trivial. The sentence says 'numerator is not 0'. This is true. The conclusion 'infinite discontinuity' is correct. I see no mathematical error. The solution is correct. Wait, let me re-read the prompt instructions. 'Find every point... classify'. The solution finds -1 and 3. Classifies them correctly. The steps are logically sound. I will mark it as pass.
  • claude-sonnet-5-5: pass 2026-10-09 — The factorization, the cancellation, the removable discontinuity at x=-1 (limit 1/2) and the infinite discontinuity at x=3 are all correct and match the stated answer. Line 5's Subs(-2, x, 3) = -2 is a trivial display, and the wording could say that the reduced numerator is nonzero, but it is not false.
  • gpt-oss:20b: pass 2026-10-09
  • qwen3.6:27b-mlx: fail (error) 2026-10-09 — Step 5 is mathematically incorrect: it substitutes x=3 into the numerator (-2x-2) to get -8, not -2. Furthermore, the logic for classifying the discontinuity at x=3 is flawed because it ignores the simplified form -2/(x-3), where the numerator is -2 (non-zero) and the denominator is zero, leading to an infinite discontinuity. The calculation shown (-2) is wrong, and the reasoning 'numerator is not 0' is applied to the wrong value.

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 2026-10-09 with SymPy 1.14.0.

← Problem 1.482All continuity and types of discontinuity problemsProblem 1.484 →