Limit of \( \displaystyle \left(2 x + 1\right) e^{- 2 x - 1} \) as \( x \to \infty \)
Problem 1.332 · medium
Evaluate \( \displaystyle \lim_{x \to \infty} \left(2 x + 1\right) e^{- 2 x - 1} \).
- \[ \lim_{x \to \infty}\left(\left(2 x + 1\right) e^{- 2 x - 1}\right) \]limit rewriteEvaluate the limit of the function as x approaches infinity. Rewrite the negative exponent using the property exp(-a) = 1/exp(a).✓ Proved
- \[ = \lim_{x \to \infty} \frac{d}{d x} \left(2 x + 1\right) e^{- 2 x - 1} \]lhopitalApply L'Hopital's rule because the limit is an indeterminate form of type infinity/infinity.✓ Proved
- \[ = \lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(2 x + 1\right)}{\frac{d}{d x} e^{2 x + 1}}\right) \]lhopitalCompute the derivatives of the numerator and the denominator.✓ Proved
- \[ = \lim_{x \to \infty} e^{- 2 x - 1} \]simplify simplify rewriteSimplify the resulting fraction. Cancel the common factor of 2. Rewrite the expression using a negative exponent.✓ Proved
- \[ = 0 \]limitThe limit of exp(-infinity) is 0.✓ 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(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 |
| 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 3 applies L'Hôpital’s rule to the derivative of the function, which is not a valid application of the rule. L'Hôpital’s rule should be applied to the original limit expression, not to its derivative. Step 4 repeats this incorrect application. The subsequent simplifications are fine, but the misuse of L'Hôpital’s rule invalidates the solution.qwen3.6:27b-mlx: fail (error) — Step 3 is incorrect because L'Hôpital's rule is an evaluation rule that transforms the limit of a function into the limit of its derivative, not a rule that inserts a 'Derivative' operator inside the limit expression. Step 4 is also defective as it applies the quotient rule for derivatives, which is not 'lhopital' (L'Hôpital's rule concerns the limit, not the differentiation mechanics), and it performs two operations (differentiating numerator and denominator) in one step.
Senior review claude-sonnet-5-5, 2026-10-06: fail — The L'Hôpital step is mis-written. Line 3 differentiates the entire fraction, and line 4 then treats that as the quotient of derivatives, so the equalities 2→3→4 hold only through coincidentally equal limit values. Line 2 should go directly to Limit(Derivative(2*x+1)/Derivative(exp(2*x+1))). The final answer 0 is correct.
gpt-oss:20b: uphold — Line 3 wraps the whole quotient in a single Derivative, which is the quotient-rule derivative of f/g rather than f'/g'. That is not L'Hôpital's rule, and the limit of (f/g)' does not equal the limit of f/g in general. Line 4 then replaces (f/g)' with f'/g', which is a different function. The two limits agree here only because both are 0.
Every verdict on record (5)
gpt-oss:20b: fail (error) 2026-10-06 — Step 3 applies L'Hôpital’s rule to the derivative of the function, which is not a valid application of the rule. L'Hôpital’s rule should be applied to the original limit expression, not to its derivative. Step 4 repeats this incorrect application. The subsequent simplifications are fine, but the misuse of L'Hôpital’s rule invalidates the solution.qwen3.6:27b-mlx: fail (error) 2026-10-06 — Step 3 is incorrect because L'Hôpital's rule is an evaluation rule that transforms the limit of a function into the limit of its derivative, not a rule that inserts a 'Derivative' operator inside the limit expression. Step 4 is also defective as it applies the quotient rule for derivatives, which is not 'lhopital' (L'Hôpital's rule concerns the limit, not the differentiation mechanics), and it performs two operations (differentiating numerator and denominator) in one step.claude-sonnet-5-5: fail (error) 2026-10-06 — The L'Hôpital step is mis-written. Line 3 differentiates the entire fraction, and line 4 then treats that as the quotient of derivatives, so the equalities 2→3→4 hold only through coincidentally equal limit values. Line 2 should go directly to Limit(Derivative(2*x+1)/Derivative(exp(2*x+1))). The final answer 0 is correct.qwen3.6:27b-mlx: pass 2026-10-06gpt-oss:20b: fail (error) 2026-10-06 — Step 3 incorrectly applies L’Hôpital’s rule by taking the derivative of the entire fraction instead of separately differentiating numerator and denominator. This step violates the rule‑application granularity requirement.
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.