Limit of \( \displaystyle \left(x + 1\right)^{2} e^{- x - 1} \) as \( x \to \infty \)
Problem 1.114 · medium
Evaluate \( \displaystyle \lim_{x \to \infty} \left(x + 1\right)^{2} e^{- x - 1} \).
- \[ \lim_{x \to \infty}\left(\left(x + 1\right)^{2} e^{- x - 1}\right) \]limit rewriteEvaluate the limit of the function as x approaches infinity. Rewrite the negative exponent using the property exp(-a) = 1/exp(a).✓ Proved
- \[ = \lim_{x \to \infty}\left(\left(x^{2} + 2 x + 1\right) e^{- x - 1}\right) \]algebraExpand the numerator.✓ Proved
- \[ = \lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(x^{2} + 2 x + 1\right)}{\frac{d}{d x} e^{x + 1}}\right) \]lhopitalApply L'Hopital's Rule because the limit is an indeterminate form of type infinity/infinity.✓ Proved
- \[ = \lim_{x \to \infty}\left(\left(2 x + 2\right) e^{- x - 1}\right) \]simplifyCompute the derivatives of the numerator and denominator.✓ Proved
- \[ = \lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(2 x + 2\right)}{\frac{d}{d x} e^{x + 1}}\right) \]lhopitalApply L'Hopital's Rule again.✓ Proved
- \[ = \lim_{x \to \infty}\left(2 e^{- x - 1}\right) \]simplifyCompute the derivatives of the numerator and denominator.✓ Proved
- \[ = \lim_{x \to \infty} 0 \]infinityAs x approaches infinity, exp(x + 1) approaches infinity.✓ Proved
- \[ = 0 \]limitThe limit is 0.✓ 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(x + 1), x) = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where Derivative(exp(x + 1), x) = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where Derivative(exp(x + 1), x) = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where Derivative(exp(x + 1), x) = 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 |
| 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: passqwen3.6:27b-mlx: pass
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-03gpt-oss:20b: pass 2026-10-03qwen3.6:27b-mlx: pass 2026-10-03 — The solution correctly applies L'Hopital's Rule twice to resolve the indeterminate form, with appropriate algebraic expansions and simplifications at each step.gpt-oss:20b: pass 2026-10-03
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 2026-10-03 with SymPy 1.14.0.