Limits using Taylor series
Problem 7.705 · medium
Use Taylor series to evaluate \( \displaystyle \lim_{x\to 0} \frac{- 3 x + 3 \ln{\left(x + 1 \right)}}{x^{2}} \).
- Near 0, -x + log(x + 1) = x**3/3 - x**2/2 + ⋯ (its Maclaurin series).
- \[ \lim_{x \to 0}\left(\frac{- \frac{x^{3}}{3} + \frac{x^{2}}{2} - x + \ln{\left(x + 1 \right)}}{x^{3}}\right) = 0 \]The polynomial matches the numerator to that order.✓ Proved
- \[ \frac{x^{3} - \frac{3 x^{2}}{2}}{x^{2}} = x - \frac{3}{2} \]Divide through by the denominator.✓ Proved
- \[ \lim_{x \to 0}\left(\frac{- 3 x + 3 \ln{\left(x + 1 \right)}}{x^{2}}\right) = - \frac{3}{2} \]Let x → 0: only the constant term survives.✓ Proved
Answer \( - \frac{3}{2} \)
Lines: 3 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 |
| 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: passqwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The Maclaurin series expansion in step 1 is incorrect; the term $-x$ cancels with the first term of $\ln(1+x)$, leaving $-x^2/2$ as the leading term,
Every verdict on record (6)
qwen3.6:27b-mlx: inconclusive 2026-10-19 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The Maclaurin series expansion in step 1 is incorrect; the term $-x$ cancels with the first term of $\ln(1+x)$, leaving $-x^2/2$ as the leading term,gpt-oss:20b: pass 2026-10-19qwen3.6:27b-mlx: fail (error) 2026-10-16 — Step 1 contains a factual error in the Maclaurin series expansion: the term -x^2/2 is missing, and x^3/3 is incorrectly included as a leading term for the expression -x + ln(1+x) (which actually starts with -x^2/2). This incorrect series leads to the wrong limit in Step 4.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (error) 2026-10-15 — The Maclaurin series expansion in step 1 is incorrect; it lists terms in descending order of magnitude ($x^3$ before $x^2$) and omits the linear term $x$, which is crucial for cancellation. The correct expansion is $x - x^2/2 + x^3/3 - \dots$, so $-x + \ln(1+x) = -x^2/2 + x^3/3 - \dots$. While the final answer is correct, the reasoning presented is mathematically invalid and misleading.gpt-oss:20b: inconclusive 2026-10-15 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"The solution’s second sentence misapplies the Taylor expansion: it incorrectly constructs a polynomial that does not match the numerator’s order and uses
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 2026-10-19 with SymPy 1.14.0.