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Home›Calculus 1›L'Hôpital's rule›Problem 1.429

Limit of \( \displaystyle \left(3 x + 1\right)^{2} e^{- 3 x - 1} \) as \( x \to \infty \)

Problem 1.429 · medium

Evaluate \( \displaystyle \lim_{x \to \infty} \left(3 x + 1\right)^{2} e^{- 3 x - 1} \).
  1. \[ \lim_{x \to \infty}\left(\left(3 x + 1\right)^{2} e^{- 3 x - 1}\right) \]
    limit rewriteStart with the original limit. Rewrite the negative exponent using division.✓ Proved
  2. \[ = \lim_{x \to \infty}\left(\left(9 x^{2} + 6 x + 1\right) e^{- 3 x - 1}\right) \]
    algebraExpand the squared binomial.✓ Proved
  3. \[ = \lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(9 x^{2} + 6 x + 1\right)}{\frac{d}{d x} e^{3 x + 1}}\right) \]
    lhopitalApply L'Hopital's rule because the limit is of the form infinity/infinity.✓ Proved
  4. \[ = \lim_{x \to \infty}\left(\frac{\left(18 x + 6\right) e^{- 3 x - 1}}{3}\right) \]
    simplifyCalculate the derivatives.✓ Proved
  5. \[ = \lim_{x \to \infty}\left(\frac{\left(18 x + 6\right) e^{- 3 x}}{3 e}\right) \]
    simplifySplit the exponential term.✓ Proved
  6. \[ = \lim_{x \to \infty}\left(\frac{\left(18 x + 6\right) e^{- 3 x}}{3}\right) \]
    simplifyRemove the constant factor exp(1).✓ Proved
  7. \[ = \lim_{x \to \infty}\left(\left(6 x + 2\right) e^{- 3 x}\right) \]
    simplifyDivide numerator and denominator by 3.✓ Proved
  8. \[ = \lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(6 x + 2\right)}{\frac{d}{d x} e^{3 x}}\right) \]
    lhopitalApply L'Hopital's rule again.✓ Proved
  9. \[ = \lim_{x \to \infty}\left(2 e^{- 3 x}\right) \]
    simplify simplifyCalculate the derivatives. Simplify the fraction.✓ Proved
  10. \[ = \lim_{x \to \infty} 0 \]
    infinityAs x approaches infinity, the denominator grows to infinity.✓ Proved
  11. \[ = 0 \]
    limitThe limit evaluates to 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. The reviewers disagree about how one step is explained; every verdict is in the receipt.

The full receipt
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0line 1 is the problem as stated
2✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
3✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
4✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
undefined where Derivative(exp(3*x + 1), x) = 0
5✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
undefined where Derivative(exp(3*x + 1), x) = 0
6✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
7✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
8✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
9✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
undefined where Derivative(exp(3*x), x) = 0
10✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
undefined where Derivative(exp(3*x), x) = 0
11✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
12✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
13✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
answer✓ Provedsympy 1.14.0final line against the stated answer: simplify(a - b) reduced to 0
answer, a second way✓ Provedsympy 1.14.0SymPy 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 limit that is not of the indeterminate form ∞/∞ (the original limit tends to 0, not ∞/∞).
  • qwen3.6:27b-mlx: pass — The solution correctly applies L'Hopital's rule twice to resolve the infinity/infinity form, with each step changing only one aspect of the expression and using valid labels from the fixed vocabulary.
Every verdict on record (4)
  • gpt-oss:20b: fail (error) 2026-10-08 — Step 4 incorrectly applies L'Hôpital’s rule to a limit that is not of the indeterminate form ∞/∞ (the original limit tends to 0, not ∞/∞).
  • qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly applies L'Hopital's rule twice to resolve the infinity/infinity form, with each step changing only one aspect of the expression and using valid labels from the fixed vocabulary.
  • qwen3.6:27b-mlx: fail (error) 2026-10-08 — Step 7 incorrectly removes the constant factor exp(1) from the denominator without adjusting the numerator, violating algebraic equality. Step 12 uses informal notation '2 / oo' instead of applying a limit law to the constant numerator and infinite denominator.
  • gpt-oss:20b: fail (error) 2026-10-08 — Step 4 incorrectly applies L'Hôpital’s rule to a limit that is not of the indeterminate form ∞/∞ (the denominator grows exponentially while the numerator grows only polynomially). The subsequent steps propagate this mistake, leading to an unjustified conclusion.

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 2026-10-08 with SymPy 1.14.0.