Limit of \( \displaystyle \left(2 x + 1\right)^{2} e^{- 2 x - 1} \) as \( x \to \infty \)
Problem 1.461 · medium
Evaluate \( \displaystyle \lim_{x \to \infty} \left(2 x + 1\right)^{2} e^{- 2 x - 1} \).
- \[ \lim_{x \to \infty}\left(\left(2 x + 1\right)^{2} e^{- 2 x - 1}\right) \]limit rewriteStart with the original limit. Rewrite the negative exponent as a fraction.✓ Proved
- \[ = \lim_{x \to \infty}\left(\left(4 x^{2} + 4 x + 1\right) e^{- 2 x - 1}\right) \]algebraExpand the squared binomial.✓ Proved
- \[ = \lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(4 x^{2} + 4 x + 1\right)}{\frac{d}{d x} e^{2 x + 1}}\right) \]lhopitalApply L'Hopital's rule because it is an indeterminate form of type infinity/infinity.✓ Proved
- \[ = \lim_{x \to \infty}\left(\frac{\left(8 x + 4\right) e^{- 2 x - 1}}{2}\right) \]simplifyCompute the derivatives.✓ Proved
- \[ = \lim_{x \to \infty}\left(\left(4 x + 2\right) e^{- 2 x - 1}\right) \]simplifyDivide the numerator and denominator by 2.✓ Proved
- \[ = \lim_{x \to \infty}\left(\frac{\left(4 x + 2\right) e^{- 2 x}}{e}\right) \]algebraSplit the exponential term.✓ Proved
- \[ = \frac{\lim_{x \to \infty}\left(\left(4 x + 2\right) e^{- 2 x}\right)}{e} \]limit-lawApply the limit law for products.✓ Proved
- \[ = \frac{\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(4 x + 2\right)}{\frac{d}{d x} e^{2 x}}\right)}{e} \]lhopitalApply L'Hopital's rule again.✓ Proved
- \[ = \frac{\lim_{x \to \infty}\left(2 e^{- 2 x}\right)}{e} \]simplify simplifyCompute the derivatives. Simplify the fraction.✓ Proved
- \[ = 0 \]limit simplifyEvaluate the remaining limit. Final result.✓ Proved
Answer \( 0 \)
✓ 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 + 1), x) = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where Derivative(exp(2*x + 1), 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 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where Derivative(exp(2*x), x) = 0 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where Derivative(exp(2*x), x) = 0 |
| 11 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 12 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 13 | ✓ 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 non‑indeterminate form (the original limit is 0/∞, not ∞/∞). This invalidates the subsequent steps that rely on that differentiation.qwen3.6:27b-mlx: pass
Senior review claude-sonnet-5-5, 2026-10-09: pass — Every step is algebraically valid and SymPy proved each line. The rule names (rewrite, algebra, lhopital, limit-law, simplify, limit) fit the steps. L'Hôpital is correctly applied twice to ∞/∞ forms, and the final value 0 is correct.
gpt-oss:20b: dismiss — The numerator 4x²+4x+1 and the denominator exp(2x+1) both tend to infinity, so the form is ∞/∞ and L'Hôpital applies; that the limit turns out to be 0 does not make the form determinate.
Every verdict on record (5)
gpt-oss:20b: fail (error) 2026-10-09 — Step 4 incorrectly applies L'Hôpital’s rule to a non‑indeterminate form (the original limit is 0/∞, not ∞/∞). This invalidates the subsequent steps that rely on that differentiation.qwen3.6:27b-mlx: pass 2026-10-09claude-sonnet-5-5: pass 2026-10-09 — Every step is algebraically valid and SymPy proved each line. The rule names (rewrite, algebra, lhopital, limit-law, simplify, limit) fit the steps. L'Hôpital is correctly applied twice to ∞/∞ forms, and the final value 0 is correct.gpt-oss:20b: fail (error) 2026-10-09 — Step 4 incorrectly applies L’Hôpital’s rule to a fraction that is not of the indeterminate form ∞/∞ or 0/0. As x→∞, the numerator (4x²+4x+1) grows polynomially while the denominator exp(2x+1) grows exponentially, so the limit is 0, not an indeterminate form. This misuse of L’Hôpital’s rule invalidates the subsequent steps.qwen3.6:27b-mlx: pass 2026-10-09 — The solution correctly applies L'Hopital's rule and limit laws, with each step changing only one aspect of the expression and using valid labels from the fixed vocabulary.
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 2026-10-09 with SymPy 1.14.0.