Limit of \( \displaystyle \frac{x + 1}{\sin{\left(x + 1 \right)}} \) as \( x \to -1 \)
Problem 1.327 · easy
Evaluate \( \displaystyle \lim_{x \to -1} \frac{x + 1}{\sin{\left(x + 1 \right)}} \).
- \[ \lim_{x \to -1^+}\left(\frac{x + 1}{\sin{\left(x + 1 \right)}}\right) \]limitEvaluate the limit of the function as x approaches -1.✓ Proved
- \[ = \lim_{x \to -1^+}\left(\frac{\sin{\left(x + 1 \right)}}{x + 1}\right) \]rewrite algebraRewrite the expression to use the reciprocal identity. Invert the fraction.✓ Proved
- \[ = \lim_{x \to -1^+}\left(\frac{\frac{d}{d x} \sin{\left(x + 1 \right)}}{\frac{d}{d x} \left(x + 1\right)}\right) \]lhopitalApply L'Hôpital's rule because the limit is of the form 0/0.✓ Proved
- \[ = \lim_{x \to -1^+} \cos{\left(x + 1 \right)} \]simplifyCompute the derivatives of the numerator and denominator.✓ Proved
- \[ = \lim_{x \to -1^+} 1 \]substitutionSubstitute x = -1 into the expression.✓ Proved
- \[ = 1 \]simplifyEvaluate the final constant value.✓ Proved
Answer \( 1 \)
Every line of this solution was proved by the computer algebra system SymPy. 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 | ✓ Proved | sympy 1.14.0 | line 1 is the problem as stated |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where sin(x + 1) = 0 undefined where x + 1 = 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where x + 1 = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where x + 1 = 0 undefined where Derivative(x + 1, x) = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where Derivative(x + 1, x) = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 |
| answer, a second way | ✓ Proved | sympy 1.14.0 | SymPy took the limit from both sides and got the stated value |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (style) — Steps 2 and 3 are redundant and inefficient; the expression in Step 1 is already in a form suitable for the standard limit or L'Hôpital's rule, making the reciprocal rewrite unnecessary. Additionally, Step 4 applies L'Hôpital's rule to the expression in Step 3, but the label 'lhopital' is acceptable; however, the transition from Step 3 to Step 4 is valid, but the prior steps were stylistically poor. More critically, Step 2 claims to use a 'reciprocal identity' but simply inverts the fraction, which is algebra, not a trigonometric identity, making the note misleading.
Senior review claude-sonnet-5-5, 2026-10-06: fail — The step from line 1 to line 2 treats lim f and lim 1/f as equal under a 'reciprocal identity'. That is false in general, and SymPy's check passes only because both limits evaluate to 1. A student would learn a false rule. The rest of the chain (L'Hôpital, derivative evaluation, substitution) is acceptable under the contract.
qwen3.6:27b-mlx: uphold — Line 2 is a real defect: it equates lim f with lim 1/f, which is false in general (the correct relation is lim f = 1/lim(1/f)) and holds here only because the value is 1; the claim about line 3 is wrong, since inverting the fraction there is valid algebra.
Every verdict on record (5)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (style) 2026-10-06 — Steps 2 and 3 are redundant and inefficient; the expression in Step 1 is already in a form suitable for the standard limit or L'Hôpital's rule, making the reciprocal rewrite unnecessary. Additionally, Step 4 applies L'Hôpital's rule to the expression in Step 3, but the label 'lhopital' is acceptable; however, the transition from Step 3 to Step 4 is valid, but the prior steps were stylistically poor. More critically, Step 2 claims to use a 'reciprocal identity' but simply inverts the fraction, which is algebra, not a trigonometric identity, making the note misleading.claude-sonnet-5-5: fail (misleading) 2026-10-06 — The step from line 1 to line 2 treats lim f and lim 1/f as equal under a 'reciprocal identity'. That is false in general, and SymPy's check passes only because both limits evaluate to 1. A student would learn a false rule. The rest of the chain (L'Hôpital, derivative evaluation, substitution) is acceptable under the contract.qwen3.6:27b-mlx: inconclusive 2026-10-06 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "Steps 2 and 3 are logically invalid; rewriting f(x) as 1/f(x) changes the limit value (from 1 to 1, coincidentally, but the operation is not an equigpt-oss:20b: pass 2026-10-06
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
gemma4:26b, checked 2026-10-06 with SymPy 1.14.0.