Limit of \( \displaystyle \left(3 x + 1\right) e^{- 3 x - 1} \) as \( x \to \infty \)
Problem 1.501 · medium
Evaluate \( \displaystyle \lim_{x \to \infty} \left(3 x + 1\right) e^{- 3 x - 1} \).
- \[ \lim_{x \to \infty}\left(\left(3 x + 1\right) e^{- 3 x - 1}\right) \]limit rewriteEvaluate the limit of the function as x approaches infinity. Rewrite the negative exponent as a fraction.✓ Proved
- \[ = \lim_{x \to \infty}\left(3 e^{- 3 x - 1}\right) \]lhopitalApply L'Hôpital's rule because the limit is of the form infinity/infinity.✓ Proved
- \[ = \lim_{x \to \infty} e^{- 3 x - 1} \]simplify simplify rewriteCompute the derivatives of the numerator and denominator. Simplify the fraction by canceling the common factor 3. Rewrite the expression using a negative exponent.✓ Proved
- \[ = 0 \]limitThe limit of exp(-u) as u approaches 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 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 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 |
| 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 changes the expression from 3/exp(3x+1) to 3/(3*exp(3x+1)), which is not algebraically equivalent. The correct simplification would directly cancel the factor 3 to obtain 1/exp(3x+1) without introducing an extra factor of 3 in the denominator.qwen3.6:27b-mlx: fail (error) — Step 3 claims to apply L'Hôpital's rule but fails to differentiate the numerator, changing (3*x + 1) to 3 without showing the derivative step. Step 4 then claims to compute derivatives, but the expression 3/exp(3*x + 1) is already the result of differentiating the numerator (constant 3) and denominator (3*exp(3*x+1)), so Step 4 is redundant or mislabeled as 'simplify' when it should be the differentiation step, or Step 3 should have included the differentiation. Specifically, Step 3 jumps from the original fraction to the ratio of derivatives, but the numerator derivative of (3x+1) is 3, and the denominator derivative of exp(3x+1) is 3*exp(3x+1). Step 3 shows 3/exp(3x+1), which is missing the factor of 3 in the denominator. Step 4 then introduces the 3 in the denominator. This splits the application of L'Hôpital's rule across two steps incorrectly or mislabels the algebraic simplification in Step 4 as 'simplify' when it's part of the L'Hôpital process. More critically, Step 3 is mathematically incorrect as a single step labeled 'lhopital' because it does not produce the correct ratio of derivatives (it misses the chain rule factor in the denominator). Step 4 fixes this but is labeled 'simplify'. A single step cannot apply L'Hôpital's rule and then partially simplify, or apply L'Hôpital's rule incorrectly. The transition from 2 to 3 is not a valid application of L'Hôpital's rule because the denominator's derivative is 3*exp(3x+1), not exp(3x+1).
Senior review claude-sonnet-5-5, 2026-10-10: fail — Line 3 is not a valid L'Hôpital result: the correct quotient of derivatives is 3/(3*exp(3x+1)), but line 3 differentiates only the numerator. Lines 3 and 4 differ by a factor of 3 as functions. SymPy accepted them only because both limits evaluate to 0, so the equality check was vacuous. Step 4 is also mislabeled 'simplify' when it applies L'Hôpital.
gpt-oss:20b: uphold — Step 4 introduces the denominator's derivative 3*exp(3x+1), which is the L'Hôpital step, so the label 'simplify' names a rule the step did not apply.qwen3.6:27b-mlx: uphold — Line 3 differentiates only the numerator and leaves the denominator as exp(3x+1), so it is not the quotient of derivatives; the denominator is differentiated only in step 4, which splits one L'Hôpital application across two steps.
Every verdict on record (5)
gpt-oss:20b: fail (error) 2026-10-10 — Step 4 changes the expression from 3/exp(3x+1) to 3/(3*exp(3x+1)), which is not algebraically equivalent. The correct simplification would directly cancel the factor 3 to obtain 1/exp(3x+1) without introducing an extra factor of 3 in the denominator.qwen3.6:27b-mlx: fail (error) 2026-10-10 — Step 3 claims to apply L'Hôpital's rule but fails to differentiate the numerator, changing (3*x + 1) to 3 without showing the derivative step. Step 4 then claims to compute derivatives, but the expression 3/exp(3*x + 1) is already the result of differentiating the numerator (constant 3) and denominator (3*exp(3*x+1)), so Step 4 is redundant or mislabeled as 'simplify' when it should be the differentiation step, or Step 3 should have included the differentiation. Specifically, Step 3 jumps from the original fraction to the ratio of derivatives, but the numerator derivative of (3x+1) is 3, and the denominator derivative of exp(3x+1) is 3*exp(3x+1). Step 3 shows 3/exp(3x+1), which is missing the factor of 3 in the denominator. Step 4 then introduces the 3 in the denominator. This splits the application of L'Hôpital's rule across two steps incorrectly or mislabels the algebraic simplification in Step 4 as 'simplify' when it's part of the L'Hôpital process. More critically, Step 3 is mathematically incorrect as a single step labeled 'lhopital' because it does not produce the correct ratio of derivatives (it misses the chain rule factor in the denominator). Step 4 fixes this but is labeled 'simplify'. A single step cannot apply L'Hôpital's rule and then partially simplify, or apply L'Hôpital's rule incorrectly. The transition from 2 to 3 is not a valid application of L'Hôpital's rule because the denominator's derivative is 3*exp(3x+1), not exp(3x+1).claude-sonnet-5-5: fail (error) 2026-10-10 — Line 3 is not a valid L'Hôpital result: the correct quotient of derivatives is 3/(3*exp(3x+1)), but line 3 differentiates only the numerator. Lines 3 and 4 differ by a factor of 3 as functions. SymPy accepted them only because both limits evaluate to 0, so the equality check was vacuous. Step 4 is also mislabeled 'simplify' when it applies L'Hôpital.qwen3.6:27b-mlx: fail (error) 2026-10-10 — Step 3 claims to apply L'Hôpital's rule but fails to differentiate the numerator, jumping directly from (3*x + 1) to 3 without showing the derivative step or applying the rule to both numerator and denominator simultaneously. Step 4 then performs the differentiation, violating the one-rule-per-step constraint by splitting the application of L'Hôpital's rule across two steps.gpt-oss:20b: fail (style) 2026-10-10 — Step 4 applies L’Hôpital’s rule (derivatives of numerator and denominator) but is labeled as "simplify". The correct rule name for that step is "lhopital".
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 2026-10-10 with SymPy 1.14.0.