Limit of \( \displaystyle \left(x + 1\right) e^{- x - 1} \) as \( x \to \infty \)
Problem 1.317 · medium
Evaluate \( \displaystyle \lim_{x \to \infty} \left(x + 1\right) e^{- x - 1} \).
- \[ \lim_{x \to \infty}\left(\left(x + 1\right) e^{- x - 1}\right) \]limit algebra rewriteEvaluate the limit of the function as x approaches infinity. Distribute the negative sign in the exponent. Rewrite the exponential with a negative exponent as a fraction.✓ Proved
- \[ = \lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(x + 1\right) e^{- x - 1}}{\frac{d}{d x} e^{x + 1}}\right) \]lhopitalApply L'Hopital's Rule because the limit is of the form infinity/infinity.✓ Proved
- \[ = \lim_{x \to \infty}\left(\left(- \left(x + 1\right) e^{x + 1} + e^{x + 1}\right) e^{- 2 x - 2} e^{- x - 1}\right) \]simplifyCompute the derivatives of the numerator and denominator.✓ Proved
- \[ = \lim_{x \to \infty}\left(\left(- \left(x + 1\right) e^{x + 1} + e^{x + 1}\right) e^{- 3 x - 3}\right) \]algebraSimplify the complex fraction.✓ Proved
- \[ = \lim_{x \to \infty}\left(- x e^{- 2 x - 2}\right) \]simplify simplify algebraFactor out exp(x + 1) from the numerator. Simplify the expression in the numerator. Apply the power rule to the exponential term.✓ Proved
- \[ = \lim_{x \to \infty}\left(- \frac{x e^{- 2 x - 2}}{2}\right) \]lhopitalApply L'Hopital's Rule again to resolve the indeterminate form.✓ Proved
- \[ = \lim_{x \to \infty}\left(- \frac{x^{2} e^{- 2 x - 2}}{2}\right) \]simplifySimplify the derivative of the denominator.✓ Proved
- \[ = 0 \]limitThe limit evaluates to 0 as the denominator grows much faster than the numerator.✓ Proved
Answer \( 0 \)
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. A reviewer's objection to the explanation stands; see the receipt.
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 |
| 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 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 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 |
| 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: inconclusive — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"Step 4 incorrectly applies L’Hôpital’s rule: it takes the derivative of the entire fraction \n\\((x+1)/e^{x+1}\\) instead of the derivative of the numeratqwen3.6:27b-mlx: fail (error) — Step 4 incorrectly applies L'Hopital's Rule by replacing the limit of the ratio with the limit of the ratio of derivatives, but the expression in Step 4 is malformed (it shows a ratio of derivatives rather than the limit of that ratio properly structured for the next step, and Step 5's derivation of the derivative of the numerator is algebraically incorrect: the derivative of (x+1)/exp(x+1) is not simply the quotient shown in step 5 without proper simplification, but more importantly, Step 4's notation `Limit(Derivative(...)/Derivative(...))` is not a standard application of the rule which should yield `Limit( (d/dx num) / (d/dx den) )`. However, the primary error is in Step 5: The derivative of the numerator `(x+1)/exp(x+1)` is calculated incorrectly or presented in a confusingly unsimplified way that doesn't match the standard quotient rule application cleanly, but actually, looking closely at Step 5, it claims to compute derivatives. The derivative of `x+1` is 1. The derivative of `exp(x+1)` is `exp(x+1)`. So the new ratio should be `1 / exp(x+1)`. Step 5 shows a complex fraction resulting from applying quotient rule to the original fraction `(x+1)/exp(x+1)`? No, L'Hopital says take derivative of top and bottom separately. Top: `x+1` -> `1`. Bottom: `exp(x+1)` -> `exp(x+1)`. So the next limit should be `Limit(1/exp(x+1), x, oo)`. Step 5 instead seems to have taken the derivative of the *entire* fraction `(x+1)/exp(x+1)` using the quotient rule, which is not what L'Hopital's rule does. L'Hopital's rule replaces `f/g` with `f'/g'`. It does not replace `f/g` with `(f/g)'`. This is a fundamental misunderstanding of L'Hopital's Rule.
Every verdict on record (4)
gpt-oss:20b: inconclusive 2026-10-06 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"Step 4 incorrectly applies L’Hôpital’s rule: it takes the derivative of the entire fraction \n\\((x+1)/e^{x+1}\\) instead of the derivative of the numeratqwen3.6:27b-mlx: fail (error) 2026-10-06 — Step 4 incorrectly applies L'Hopital's Rule by replacing the limit of the ratio with the limit of the ratio of derivatives, but the expression in Step 4 is malformed (it shows a ratio of derivatives rather than the limit of that ratio properly structured for the next step, and Step 5's derivation of the derivative of the numerator is algebraically incorrect: the derivative of (x+1)/exp(x+1) is not simply the quotient shown in step 5 without proper simplification, but more importantly, Step 4's notation `Limit(Derivative(...)/Derivative(...))` is not a standard application of the rule which should yield `Limit( (d/dx num) / (d/dx den) )`. However, the primary error is in Step 5: The derivative of the numerator `(x+1)/exp(x+1)` is calculated incorrectly or presented in a confusingly unsimplified way that doesn't match the standard quotient rule application cleanly, but actually, looking closely at Step 5, it claims to compute derivatives. The derivative of `x+1` is 1. The derivative of `exp(x+1)` is `exp(x+1)`. So the new ratio should be `1 / exp(x+1)`. Step 5 shows a complex fraction resulting from applying quotient rule to the original fraction `(x+1)/exp(x+1)`? No, L'Hopital says take derivative of top and bottom separately. Top: `x+1` -> `1`. Bottom: `exp(x+1)` -> `exp(x+1)`. So the next limit should be `Limit(1/exp(x+1), x, oo)`. Step 5 instead seems to have taken the derivative of the *entire* fraction `(x+1)/exp(x+1)` using the quotient rule, which is not what L'Hopital's rule does. L'Hopital's rule replaces `f/g` with `f'/g'`. It does not replace `f/g` with `(f/g)'`. This is a fundamental misunderstanding of L'Hopital's Rule.gpt-oss:20b: inconclusive 2026-10-06 — reviewer response could not be parsed: {"verdict":"fail","severity":"misleading","notes":"Step 10 applies L'Hôpital’s rule to the limit \n\[-x/\exp(2x+2)\], which is not an indeterminate form (the numerator tends to \(-\infty\) while the dqwen3.6:27b-mlx: fail (error) 2026-10-06 — Step 4 misapplies L'Hopital's Rule by differentiating the denominator as if it were just exp(x+1) instead of the full denominator exp(x+1) from Step 3, and Step 5 computes the derivative of the numerator incorrectly (missing the quotient rule structure or product rule application consistent with the previous step). Specifically, Step 4 claims to apply L'Hopital's but the resulting expression in Step 5 does not match the derivative of the fraction in Step 3 divided by the derivative of its denominator. The transition from Step 3 to Step 4/5 is mathematically incoherent.
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 2026-10-06 with SymPy 1.14.0.