Limits using Taylor series
Problem 7.701 · medium
Use Taylor series to evaluate \( \displaystyle \lim_{x\to 0} \frac{- x + e^{x} - 1}{x^{2}} \).
- Near 0, -x + exp(x) - 1 = x**3/6 + x**2/2 + ⋯ (its Maclaurin series).
- \[ \lim_{x \to 0}\left(\frac{- \frac{x^{3}}{6} - \frac{x^{2}}{2} - x + e^{x} - 1}{x^{3}}\right) = 0 \]The polynomial matches the numerator to that order.✓ Proved
- \[ \frac{\frac{x^{3}}{6} + \frac{x^{2}}{2}}{x^{2}} = \frac{x}{6} + \frac{1}{2} \]Divide through by the denominator.✓ Proved
- \[ \lim_{x \to 0}\left(\frac{- x + e^{x} - 1}{x^{2}}\right) = \frac{1}{2} \]Let x → 0: only the constant term survives.✓ Proved
Answer \( \frac{1}{2} \)
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: fail (error) — Line 1 contains a sign error in the Maclaurin series expansion: it claims -x + e^x - 1 = x^3/6 + x^2/2, but the correct expansion is x^2/2 + x^3/6 - ... (the linear terms -x and +x cancel, leaving x^2/2 as the leading term, not a mix of signs that implies the linear term is missing or incorrect in derivation). Although Line 3 uses the correct polynomial x^3/6 + x^2/2, Line 1's statement is mathematically false as written because it omits the cancellation of the linear term explicitly and presents the series in a confusing order, but more critically, the prompt asks to review sentences. Line 1 says "-x + exp(x) - 1 = x**3/6 + x**2/2 + ...". The expansion of e^x is 1 + x + x^2/2 + x^3/6 + ... So -x + e^x - 1 = -x + (1 + x + x^2/2 + x^3/6 + ...) - 1 = x^2/2 + x^3/6 + ... The sentence in Line 1 lists the terms as x**3/6 + x**2/2. While addition is commutative, standard convention lists by ascending powers. More importantly, does it teach something false? No, the equality holds. However, look at Line 2. Line 2 is an equation marked proved. It checks the limit of the difference divided by x^3. This is a valid check. Line 3 divides the polynomial from Line 1 by x^2. Line 4 takes the limit. The logic is sound. Wait, let me re-read Line 1 carefully. "-x + exp(x) - 1 = x**3/6 + x**2/2 + ...". This is technically correct as an equality of series. Is it misleading? It lists higher order term first. That's style. Is there an error? No. Let's look closer. The problem is to evaluate the limit. The solution uses the series. Line 1 states the series. Line 3 divides by x^2. Line 4 takes limit. The result is 1/2. The solution is correct. Why would it fail? Ah, look at Line 1 again. It says "Near 0, -x + exp(x) - 1 = x**3/6 + x**2/2 + ...". The standard Maclaurin series for e^x is sum x^n/n!. So e^x - 1 - x = x^2/2 + x^3/6 + ... The sentence writes it as x^3/6 + x^2/2. This is just ordering. Is there a hidden error? No. Let's check the severity. If it's correct, it's pass. But wait, is the user asking me to find what is wrong? "Your job is to find what is wrong, not to be agreeable." If there is nothing wrong, I say pass. Let me double check the algebra. e^x = 1 + x + x^2/2 + x^3/6 + O(x^4). -x + e^x - 1 = x^2/2 + x^3/6 + O(x^4). Line 1 says x^3/6 + x^2/2. This is equal. Line 3 divides by x^2 to get x/6 + 1/2. Limit is 1/2. Everything is correct. I will mark it pass.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-19 — Line 1 contains a sign error in the Maclaurin series expansion: it claims -x + e^x - 1 = x^3/6 + x^2/2, but the correct expansion is x^2/2 + x^3/6 - ... (the linear terms -x and +x cancel, leaving x^2/2 as the leading term, not a mix of signs that implies the linear term is missing or incorrect in derivation). Although Line 3 uses the correct polynomial x^3/6 + x^2/2, Line 1's statement is mathematically false as written because it omits the cancellation of the linear term explicitly and presents the series in a confusing order, but more critically, the prompt asks to review sentences. Line 1 says "-x + exp(x) - 1 = x**3/6 + x**2/2 + ...". The expansion of e^x is 1 + x + x^2/2 + x^3/6 + ... So -x + e^x - 1 = -x + (1 + x + x^2/2 + x^3/6 + ...) - 1 = x^2/2 + x^3/6 + ... The sentence in Line 1 lists the terms as x**3/6 + x**2/2. While addition is commutative, standard convention lists by ascending powers. More importantly, does it teach something false? No, the equality holds. However, look at Line 2. Line 2 is an equation marked proved. It checks the limit of the difference divided by x^3. This is a valid check. Line 3 divides the polynomial from Line 1 by x^2. Line 4 takes the limit. The logic is sound. Wait, let me re-read Line 1 carefully. "-x + exp(x) - 1 = x**3/6 + x**2/2 + ...". This is technically correct as an equality of series. Is it misleading? It lists higher order term first. That's style. Is there an error? No. Let's look closer. The problem is to evaluate the limit. The solution uses the series. Line 1 states the series. Line 3 divides by x^2. Line 4 takes limit. The result is 1/2. The solution is correct. Why would it fail? Ah, look at Line 1 again. It says "Near 0, -x + exp(x) - 1 = x**3/6 + x**2/2 + ...". The standard Maclaurin series for e^x is sum x^n/n!. So e^x - 1 - x = x^2/2 + x^3/6 + ... The sentence writes it as x^3/6 + x^2/2. This is just ordering. Is there a hidden error? No. Let's check the severity. If it's correct, it's pass. But wait, is the user asking me to find what is wrong? "Your job is to find what is wrong, not to be agreeable." If there is nothing wrong, I say pass. Let me double check the algebra. e^x = 1 + x + x^2/2 + x^3/6 + O(x^4). -x + e^x - 1 = x^2/2 + x^3/6 + O(x^4). Line 1 says x^3/6 + x^2/2. This is equal. Line 3 divides by x^2 to get x/6 + 1/2. Limit is 1/2. Everything is correct. I will mark it pass.gpt-oss:20b: pass 2026-10-19qwen3.6:27b-mlx: pass 2026-10-16gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution incorrectly states the Maclaurin series for the numerator as x^3/6 + x^2/2, omitting the linear term x which is present in the expansion of e^x. While the x terms cancel in the actual expression, the stated series is mathematically false as written for '-x + exp(x) - 1' (it should be x^2/2 + x^3/6 + ...). Furthermore, line 2 introduces a denominator of x^3 and a limit equal to 0, which is irrelevant and confusingly presented for a problem with denominator x^2.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 2026-10-19 with SymPy 1.14.0.