Continuity and types of discontinuity
Problem 1.536 · hard
Find every point where \( \displaystyle f(x) = \frac{x - 3}{x^{2} - 9} \) 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.
- \[ x^{2} - 9 = \left(x - 3\right) \left(x + 3\right) \]Factor the denominator.✓ Proved
- \[ \frac{x - 3}{x^{2} - 9} = \frac{1}{x + 3} \]The factor (x − 3) cancels.✓ Proved
- \[ \lim_{x \to 3^+}\left(\frac{x - 3}{x^{2} - 9}\right) = \frac{1}{6} \]At x = 3 the limit exists, but f(3) is undefined: a removable discontinuity.✓ Proved
- \[ \left. 1 \right|_{\substack{ x=-3 }} = 1 \]At x = -3 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved
Answer \( \text{removable at } x = 3;\ \text{infinite at } x = -3 \)
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.
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', 3: 'removable'} |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — Line 5 is mathematically incorrect and misleading. It substitutes x=1 into the simplified expression 1/(x+3) to get 1, which is irrelevant to the behavior at x=-3. The correct analysis for x=-3 requires evaluating the limit of 1/(x+3) as x approaches -3, which is infinite, not substituting an arbitrary value.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-11 — Line 5 is mathematically incorrect and misleading. It substitutes x=1 into the simplified expression 1/(x+3) to get 1, which is irrelevant to the behavior at x=-3. The correct analysis for x=-3 requires evaluating the limit of 1/(x+3) as x approaches -3, which is infinite, not substituting an arbitrary value.gpt-oss:20b: pass 2026-10-11gpt-oss:20b: fail (misleading) 2026-10-11 — The solution incorrectly evaluates the function at x = -3 by writing Subs(1, x, -3) = 1, which is meaningless and would mislead a student into thinking the function has a finite value there. The correct approach is to note that as x→−3, 1/(x+3) → ±∞, indicating an infinite discontinuity.qwen3.6:27b-mlx: fail (error) 2026-10-11 — Step 5 is mathematically incorrect: substituting x = -3 into the simplified expression 1/(x+3) yields division by zero, not 1. The conclusion of an infinite discontinuity is correct, but the justification provided is false and would teach a student that 1/0 = 1.
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 2026-10-11 with SymPy 1.14.0.