Limit of \( \displaystyle \left(2 x - 1\right) e^{1 - 2 x} \) as \( x \to \infty \)
Problem 1.254 · medium
- \[ \lim_{x \to \infty}\left(\left(2 x - 1\right) e^{1 - 2 x}\right) \]limitStart with the original limit expression.✓ Proved
- \[ = \lim_{x \to \infty}\left(e \left(2 x - 1\right) e^{- 2 x}\right) \]rewrite algebraRewrite the exponential term using properties of exponents. Rearrange the terms for clarity.✓ Proved
- \[ = \lim_{x \to \infty}\left(\frac{\frac{d}{d x} e \left(2 x - 1\right)}{\frac{d}{d x} e^{2 x}}\right) \]lhopitalApply L'Hopital's rule because the limit is of the form infinity/infinity.✓ Proved
- \[ = \lim_{x \to \infty}\left(e e^{- 2 x}\right) \]simplify simplifyCompute the derivatives of the numerator and denominator. Cancel the common factor of 2.✓ Proved
- \[ = \lim_{x \to \infty} e^{1 - 2 x} \]algebraCombine the exponential terms.✓ Proved
- \[ = 0 \]limitAs x approaches infinity, the exponential term approaches zero.✓ Proved
✓ Nihil obstat 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. Reviewers found nothing wrong with the explanation.
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 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where Derivative(exp(2*x), x) = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where Derivative(exp(2*x), 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 |
| 8 | ✓ 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: fail (error) — Step 4 incorrectly applies L'Hôpital’s rule to a product that is not of the indeterminate form 0/0 or ∞/∞. The expression (2*x-1)/exp(2*x)*exp(1) is a constant times a quotient that tends to 0, so no differentiation is needed. The derivative step is mathematically invalid.qwen3.6:27b-mlx: pass
Senior review claude-sonnet-5-5, 2026-10-05: pass — Every line is a correct equality and SymPy proved each one. L'Hôpital is applied legitimately to ((2x-1)e)/exp(2x), a oo/oo form. The labels 'simplify' on the derivative-evaluation and factor-cancelling steps are acceptable, though 'cancel' would fit step 6 slightly better.
gpt-oss:20b: dismiss — Differentiating the numerator (2x-1)*e is valid because e is a constant factor, and the ratio of derivatives equals the correct L'Hôpital quotient for the form oo/oo; SymPy proved the line equal.
Every verdict on record (5)
qwen3.6:27b-mlx: pass 2026-10-05gpt-oss:20b: fail (error) 2026-10-05 — Step 4 incorrectly applies L'Hôpital’s rule to a product that is not of the indeterminate form 0/0 or ∞/∞. The expression (2*x-1)/exp(2*x)*exp(1) is a constant times a quotient that tends to 0, so no differentiation is needed. The derivative step is mathematically invalid.claude-sonnet-5-5: pass 2026-10-05 — Every line is a correct equality and SymPy proved each one. L'Hôpital is applied legitimately to ((2x-1)e)/exp(2x), a oo/oo form. The labels 'simplify' on the derivative-evaluation and factor-cancelling steps are acceptable, though 'cancel' would fit step 6 slightly better.qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly applies L'Hopital's rule after rewriting the expression into a quotient form. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.gpt-oss:20b: inconclusive 2026-10-05 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"Step 4 incorrectly applies L’Hôpital’s rule: it differentiates the product \((2x-1)\exp(1)\) instead of just the numerator \((2x-1)\) of the fraction \((2
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 2026-10-05 with SymPy 1.14.0.