Home›Calculus 1›Continuity and types of discontinuity›Problem 1.754

Continuity and types of discontinuity

Problem 1.754 · hard

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

Answer \( \text{removable at } x = -1;\ \text{infinite at } x = -2 \)

Lines: 4 proved, 1 reviewed. 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. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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: {-2: 'infinite', -1: 'removable'}

Reviewers

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

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-25
gpt-oss:20b: pass 2026-10-25
qwen3.6:27b-mlx: fail (error) 2026-10-24 — Line 4 only computes the right-hand limit, but for a removable discontinuity, the two-sided limit must exist (left and right limits must be equal). While true in this case, the reasoning is incomplete and technically incorrect as stated. Line 5 uses substitution on the simplified function to determine the numerator's value, which is valid, but the phrasing 'Subs... = 1' is an algebraic fact, not a limit evaluation, making the logical jump to 'infinite discontinuity' slightly abrupt without explicitly stating the limit of the simplified form is non-zero.
gpt-oss:20b: pass 2026-10-24
qwen3.6:27b-mlx: pass 2026-10-24 — The solution correctly identifies the discontinuities and classifies them based on the behavior of the simplified function and limits.
gpt-oss:20b: pass 2026-10-24

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 2026-10-25 with SymPy 1.14.0.