Limits using Taylor series
Problem 7.608 · medium
Use Taylor series to evaluate \( \displaystyle \lim_{x\to 0} \frac{2 - 2 \cos{\left(x \right)}}{x \sin{\left(x \right)}} \).
- Near 0, 1 - cos(x) = -x**4/24 + x**2/2 + ⋯ (its Maclaurin series).
- \[ \lim_{x \to 0}\left(\frac{\frac{x^{4}}{24} - \frac{x^{2}}{2} - \cos{\left(x \right)} + 1}{x^{4}}\right) = 0 \]The polynomial matches the numerator to that order.✓ Proved
- \[ \lim_{x \to 0}\left(\frac{2 - 2 \cos{\left(x \right)}}{x \sin{\left(x \right)}}\right) = 1 \]Let x → 0: only the constant term survives.✓ Proved
Answer \( 1 \)
Lines: 2 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. How we verify
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 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the expression at x = ±10⁻⁸ in 50-digit arithmetic |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution claims the limit is 1, but the correct value is 1/2. The reasoning in step 3 is flawed: substituting the series 1 - cos(x) ~ x^2/2 and sin(x) ~ x yields (2 * x^2/2) / (x * x) = 1, which is incorrect because the numerator is 2(1-cos(x)) ~ x^2, so the ratio is x^2/x^2 = 1? Wait. 1-cos(x) = x^2/2 - x^4/24. So 2(1-cos(x)) = x^2 - x^4/12. Denominator x*sin(x) = x(x - x^3/6) = x^2 - x^4/6. Limit is (x^2)/(x^2) = 1. Actually, the limit IS 1. Let me re-evaluate. Limit x->0 of (2 - 2cos(x))/(x sin(x)). Numerator: 2(1 - (1 - x^2/2 + x^4/24 - ...)) = 2(x^2/2 - x^4/24 + ...) = x^2 - x^4/12 + ... Denominator: x(x - x^3/6 + ...) = x^2 - x^4/6 + ... Ratio: (x^2 - x^4/12) / (x^2 - x^4/6) = (1 - x^2/12) / (1 - x^2/6) -> 1/1 = 1. Okay, the answer 1 is correct. Why did I think it was 1/2? L'Hopital's rule: Deriv Num: 2sin(x) Deriv Denom: sin(x) + xcos(x) Limit 2sin(x)/(sin(x)+xcos(x)) -> 0/0. Apply L'Hopital again: Deriv Num: 2cos(x) Deriv Denom: cos(x) + cos(x) - xsin(x) = 2cos(x) - xsin(x) Limit 2cos(0)/(2cos(0) - 0) = 2/2 = 1. So the answer is indeed 1. The solution steps are: 1. States series for 1-cos(x). 2. Checks a limit involving x^4, which seems like a distractor or a check on higher order terms, but the comment says 'polynomial matches numerator to that order'. 3. Concludes limit is 1. Is there an error in the sentences? Sentence 1: 'Near 0, 1 - cos(x) = -x**4/24 + x**2/2 + ...'. This is correct. Sentence 3: 'Let x -> 0: only the constant term survives.' This is a bit vague but in the context of dividing the leading terms x^2/x^2, the constant term of the ratio is 1. Wait, look at step 2. It calculates Limit((x**4/24 - x**2/2 - cos(x) + 1)/x**4, x, 0, dir='+-') = 0. Numerator of this limit: x^4/24 - x^2/2 + 1 - cos(x). Since 1 - cos(x) = x^2/2 - x^4/24 + O(x^6), Num = x^4/24 - x^2/2 + x^2/2 - x^4/24 + O(x^6) = O(x^6). Limit O(x^6)/x^4 = 0. This is correct. The solution seems correct. The verdict should be pass. Let me double check the 'adversarial' nature. Is there a subtle error? Sentence 1 writes the series as `-x**4/24 + x**2/2`. Usually we write `x^2/2 - x^4/24`. The order is reversed but it's a sum, so it's fine. Sentence 3 says 'only the constant term survives'. This is standard shorthand for 'the limit is the ratio of the leading coefficients'. I will mark it as pass.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The solution claims the limit is 1, but the correct value is 1/2. The reasoning in step 3 is flawed: substituting the series 1 - cos(x) ~ x^2/2 and sin(x) ~ x yields (2 * x^2/2) / (x * x) = 1, which is incorrect because the numerator is 2(1-cos(x)) ~ x^2, so the ratio is x^2/x^2 = 1? Wait. 1-cos(x) = x^2/2 - x^4/24. So 2(1-cos(x)) = x^2 - x^4/12. Denominator x*sin(x) = x(x - x^3/6) = x^2 - x^4/6. Limit is (x^2)/(x^2) = 1. Actually, the limit IS 1. Let me re-evaluate. Limit x->0 of (2 - 2cos(x))/(x sin(x)). Numerator: 2(1 - (1 - x^2/2 + x^4/24 - ...)) = 2(x^2/2 - x^4/24 + ...) = x^2 - x^4/12 + ... Denominator: x(x - x^3/6 + ...) = x^2 - x^4/6 + ... Ratio: (x^2 - x^4/12) / (x^2 - x^4/6) = (1 - x^2/12) / (1 - x^2/6) -> 1/1 = 1. Okay, the answer 1 is correct. Why did I think it was 1/2? L'Hopital's rule: Deriv Num: 2sin(x) Deriv Denom: sin(x) + xcos(x) Limit 2sin(x)/(sin(x)+xcos(x)) -> 0/0. Apply L'Hopital again: Deriv Num: 2cos(x) Deriv Denom: cos(x) + cos(x) - xsin(x) = 2cos(x) - xsin(x) Limit 2cos(0)/(2cos(0) - 0) = 2/2 = 1. So the answer is indeed 1. The solution steps are: 1. States series for 1-cos(x). 2. Checks a limit involving x^4, which seems like a distractor or a check on higher order terms, but the comment says 'polynomial matches numerator to that order'. 3. Concludes limit is 1. Is there an error in the sentences? Sentence 1: 'Near 0, 1 - cos(x) = -x**4/24 + x**2/2 + ...'. This is correct. Sentence 3: 'Let x -> 0: only the constant term survives.' This is a bit vague but in the context of dividing the leading terms x^2/x^2, the constant term of the ratio is 1. Wait, look at step 2. It calculates Limit((x**4/24 - x**2/2 - cos(x) + 1)/x**4, x, 0, dir='+-') = 0. Numerator of this limit: x^4/24 - x^2/2 + 1 - cos(x). Since 1 - cos(x) = x^2/2 - x^4/24 + O(x^6), Num = x^4/24 - x^2/2 + x^2/2 - x^4/24 + O(x^6) = O(x^6). Limit O(x^6)/x^4 = 0. This is correct. The solution seems correct. The verdict should be pass. Let me double check the 'adversarial' nature. Is there a subtle error? Sentence 1 writes the series as `-x**4/24 + x**2/2`. Usually we write `x^2/2 - x^4/24`. The order is reversed but it's a sum, so it's fine. Sentence 3 says 'only the constant term survives'. This is standard shorthand for 'the limit is the ratio of the leading coefficients'. I will mark it as pass.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: fail (misleading) 2026-10-15 — The solution claims the limit is 1, but the correct value is 1/2. The reasoning in step 3 is flawed: dividing the leading term of the numerator (x^2) by the leading term of the denominator (x^2) yields 1/2, not 1. The text 'only the constant term survives' is mathematically nonsensical in this context.gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution incorrectly claims that 'only the constant term survives' when dividing by x*sin(x). Since x*sin(x) is O(x^2), the x^2 term in the numerator (which is not constant) determines the limit. The correct logic requires dividing the O(x^2) numerator by the O(x^2) denominator.gpt-oss:20b: fail (misleading) 2026-10-14 — The explanation in sentence 3 incorrectly states that only the constant term survives; the correct reasoning uses the leading terms of the Taylor expansions of numerator and denominator.
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/taylor_limit, checked 2026-10-17 with SymPy 1.14.0.