Continuity and types of discontinuity
Problem 1.394 · hard
Find every point where \( \displaystyle f(x) = \frac{- 5 x - 5}{x^{2} + 4 x + 3} \) 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.
- \[ x^{2} + 4 x + 3 = \left(x + 1\right) \left(x + 3\right) \]Factor the denominator.✓ Proved
- \[ \frac{- 5 x - 5}{x^{2} + 4 x + 3} = - \frac{5}{x + 3} \]The factor (x + 1) cancels.✓ Proved
- \[ \lim_{x \to -1^+}\left(\frac{- 5 x - 5}{x^{2} + 4 x + 3}\right) = - \frac{5}{2} \]At x = -1 the limit exists, but f(-1) is undefined: a removable discontinuity.✓ Proved
- \[ \left. -5 \right|_{\substack{ x=-3 }} = -5 \]At x = -3 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved
Answer \( \text{removable at } x = -1;\ \text{infinite at } x = -3 \)
Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see the receipt.
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: {-3: 'infinite', -1: 'removable'} |
Reviewers
gpt-oss:20b: fail (misleading) — Step 5 incorrectly states that the numerator evaluates to –5 at x = –3; the correct value is 10. This misstatement could mislead a student about how to check for an infinite discontinuity.qwen3.6:27b-mlx: fail (error) — Line 5 is mathematically incorrect and misleading. It claims that substituting x=-3 into the numerator (-5) results in -5, but then concludes there is an infinite discontinuity because the denominator is zero. While the conclusion (infinite discontinuity) is correct, the reasoning presented in the sentence is flawed: it implies that the non-zero numerator alone dictates the behavior without explicitly referencing the zero denominator in the limit context, and the equation shown `Subs(-5, x, -3) = -5` is trivial and does not represent the limit evaluation or the ratio behavior. More critically, the sentence says "the numerator is not 0 while the denominator is", which is the correct heuristic, but the equation provided `Subs(-5, x, -3) = -5` is just evaluating the constant numerator, which is irrelevant on its own. The step is confusingly presented. However, the bigger issue is that the solution fails to check the limit at x=-3 properly. It just states the numerator is non-zero. A rigorous check would involve showing the limit is infinity. The sentence is technically correct in its heuristic, but the equation is useless. Let's look closer. Is it an error? The sentence says "At x = -3 the numerator is not 0 while the denominator is, so |f(x)| grows without bound". This is a standard correct argument for infinite discontinuity. The equation `Subs(-5, x, -3) = -5` is just showing the numerator value. It's weak but not strictly "wrong" in a way that teaches falsehood, unless one interprets it as the value of the function. But wait, the prompt asks to find what is wrong. Line 3 cancels (x+1). This is valid for x != -1. Line 4 checks limit at -1. Line 5 checks x=-3. The argument for x=-3 is: numerator is -5 (non-zero), denominator is 0. Therefore infinite discontinuity. This is correct. Why would it fail? Perhaps because it doesn't explicitly state the limit is infinity? Or is there a subtle error? Ah, look at Line 5's equation: `Subs(-5, x, -3) = -5`. This is evaluating the simplified numerator `-5` at `x=-3`. This is correct. The sentence says "numerator is not 0 while the denominator is". This is correct. Is there a missing check? No. Is the classification correct? Yes. Is the setup correct? Yes. Why fail? Maybe the sentence in line 5 is considered "misleading" because it doesn't show the limit? Or is it an error? Let's re-read carefully. "Subs(-5, x, -3) = -5". The numerator of the simplified function is -5. The original numerator is -5x-5. At x=-3, original numerator is 10. Wait. Line 3 simplifies f(x) to -5/(x+3) for x != -1. At x=-3, the simplified form has numerator -5. The original form has numerator -5(-3)-5 = 10. The sentence says "the numerator is not 0". It doesn't specify which numerator. If it refers to the simplified numerator, it's -5. If it refers to the original, it's 10. Both are non-zero. So the logic holds. However, the equation `Subs(-5, x, -3) = -5` is evaluating the constant -5. This is trivial. Is it an error? No. Is it misleading? Maybe. But usually, these adversarial reviews look for actual mistakes. Let's look at Line 4. `Limit(..., x, -1, dir='+') = -5/2`. It only checks the right-hand limit. For a removable discontinuity, the two-sided limit must exist. Since the simplified function -5/(x+3) is continuous at x=-1, the two-sided limit exists and equals -5/2. Checking only the right-hand limit is insufficient to prove the two-sided limit exists, although it is true. This is a gap in rigor. Does it teach something false? It might teach that checking one side is enough. That is misleading. Also, Line 5's equation is weird. But the one-sided limit issue in Line 4 is a stronger candidate for "misleading" or "error" in a calculus context where two-sided limits are required for continuity/removability. The sentence in Line 4 says "the limit exists", but the equation only proves the right-hand limit exists. This is a logical gap. I will flag this as misleading/error.
Every verdict on record (4)
gpt-oss:20b: fail (misleading) 2026-10-07 — Step 5 incorrectly states that the numerator evaluates to –5 at x = –3; the correct value is 10. This misstatement could mislead a student about how to check for an infinite discontinuity.qwen3.6:27b-mlx: fail (error) 2026-10-07 — Line 5 is mathematically incorrect and misleading. It claims that substituting x=-3 into the numerator (-5) results in -5, but then concludes there is an infinite discontinuity because the denominator is zero. While the conclusion (infinite discontinuity) is correct, the reasoning presented in the sentence is flawed: it implies that the non-zero numerator alone dictates the behavior without explicitly referencing the zero denominator in the limit context, and the equation shown `Subs(-5, x, -3) = -5` is trivial and does not represent the limit evaluation or the ratio behavior. More critically, the sentence says "the numerator is not 0 while the denominator is", which is the correct heuristic, but the equation provided `Subs(-5, x, -3) = -5` is just evaluating the constant numerator, which is irrelevant on its own. The step is confusingly presented. However, the bigger issue is that the solution fails to check the limit at x=-3 properly. It just states the numerator is non-zero. A rigorous check would involve showing the limit is infinity. The sentence is technically correct in its heuristic, but the equation is useless. Let's look closer. Is it an error? The sentence says "At x = -3 the numerator is not 0 while the denominator is, so |f(x)| grows without bound". This is a standard correct argument for infinite discontinuity. The equation `Subs(-5, x, -3) = -5` is just showing the numerator value. It's weak but not strictly "wrong" in a way that teaches falsehood, unless one interprets it as the value of the function. But wait, the prompt asks to find what is wrong. Line 3 cancels (x+1). This is valid for x != -1. Line 4 checks limit at -1. Line 5 checks x=-3. The argument for x=-3 is: numerator is -5 (non-zero), denominator is 0. Therefore infinite discontinuity. This is correct. Why would it fail? Perhaps because it doesn't explicitly state the limit is infinity? Or is there a subtle error? Ah, look at Line 5's equation: `Subs(-5, x, -3) = -5`. This is evaluating the simplified numerator `-5` at `x=-3`. This is correct. The sentence says "numerator is not 0 while the denominator is". This is correct. Is there a missing check? No. Is the classification correct? Yes. Is the setup correct? Yes. Why fail? Maybe the sentence in line 5 is considered "misleading" because it doesn't show the limit? Or is it an error? Let's re-read carefully. "Subs(-5, x, -3) = -5". The numerator of the simplified function is -5. The original numerator is -5x-5. At x=-3, original numerator is 10. Wait. Line 3 simplifies f(x) to -5/(x+3) for x != -1. At x=-3, the simplified form has numerator -5. The original form has numerator -5(-3)-5 = 10. The sentence says "the numerator is not 0". It doesn't specify which numerator. If it refers to the simplified numerator, it's -5. If it refers to the original, it's 10. Both are non-zero. So the logic holds. However, the equation `Subs(-5, x, -3) = -5` is evaluating the constant -5. This is trivial. Is it an error? No. Is it misleading? Maybe. But usually, these adversarial reviews look for actual mistakes. Let's look at Line 4. `Limit(..., x, -1, dir='+') = -5/2`. It only checks the right-hand limit. For a removable discontinuity, the two-sided limit must exist. Since the simplified function -5/(x+3) is continuous at x=-1, the two-sided limit exists and equals -5/2. Checking only the right-hand limit is insufficient to prove the two-sided limit exists, although it is true. This is a gap in rigor. Does it teach something false? It might teach that checking one side is enough. That is misleading. Also, Line 5's equation is weird. But the one-sided limit issue in Line 4 is a stronger candidate for "misleading" or "error" in a calculus context where two-sided limits are required for continuity/removability. The sentence in Line 4 says "the limit exists", but the equation only proves the right-hand limit exists. This is a logical gap. I will flag this as misleading/error.gpt-oss:20b: fail (error) 2026-10-07 — The numerator at x = -3 is -5(-3)-5 = -20, not -5. The solution incorrectly evaluates the numerator, but the conclusion about an infinite discontinuity at x = -3 remains correct.qwen3.6:27b-mlx: fail (error) 2026-10-07 — Step 5 incorrectly evaluates the numerator at x = -3 as -5 (it is 10) and uses a substitution syntax that implies the limit is -5, contradicting the conclusion of an infinite discontinuity. The logic for classifying the discontinuity at x = -3 is flawed despite the final label being correct.
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.