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Limits using Taylor series

Problem 7.674 · medium

Use Taylor series to evaluate \( \displaystyle \lim_{x\to 0} \frac{x - \operatorname{atan}{\left(x \right)}}{x^{3}} \).

Near 0, x - atan(x) = -x**5/5 + x**3/3 + ⋯ (its Maclaurin series).
Reviewed
\[ \lim_{x \to 0}\left(\frac{\frac{x^{5}}{5} - \frac{x^{3}}{3} + x - \operatorname{atan}{\left(x \right)}}{x^{5}}\right) = 0 \]
The polynomial matches the numerator to that order.✓ Proved
\[ \frac{- \frac{x^{5}}{5} + \frac{x^{3}}{3}}{x^{3}} = \frac{1}{3} - \frac{x^{2}}{5} \]
Divide through by the denominator.✓ Proved
\[ \lim_{x \to 0}\left(\frac{x - \operatorname{atan}{\left(x \right)}}{x^{3}}\right) = \frac{1}{3} \]
Let x → 0: only the constant term survives.✓ Proved

Answer \( \frac{1}{3} \)

Lines: 3 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
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: pass
qwen3.6:27b-mlx: pass — The solution correctly identifies the Maclaurin series expansion and applies limit laws properly to evaluate the limit.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies the Maclaurin series expansion and applies limit laws properly to evaluate the limit.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the Maclaurin series expansion, performs the algebraic simplification, and evaluates the limit. The reasoning is sound and the steps are logically connected.
gpt-oss:20b: fail (error) 2026-10-16 — Sentence 2 incorrectly claims that the difference between the Maclaurin polynomial and the function has a zero limit when divided by x^5. In fact, x−atan(x)=x^3/3−x^5/5+O(x^7), so subtracting the polynomial x^5/5−x^3/3 yields a term of order x^3, not x^5, and the limit is not zero. The conclusion that the limit equals 1/3 is therefore not justified by the steps shown.
qwen3.6:27b-mlx: inconclusive 2026-10-15 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution claims the limit is 1/3, but the correct limit is -1/3. The Maclaurin series for atan(x) is x - x^3/3 + x^5/5 - ..., so x - atan(x) = x
gpt-oss:20b: pass 2026-10-15

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 2026-10-18 with SymPy 1.14.0.