Limit of \( \displaystyle \frac{6 x + 1}{9 x^{2} + 3} \) as \( x \to \infty \)
Problem 1.223 · medium
Evaluate \( \displaystyle \lim_{x \to \infty} \frac{6 x + 1}{9 x^{2} + 3} \).
- \[ \lim_{x \to \infty}\left(\frac{6 x + 1}{9 x^{2} + 3}\right) \]limitStart with the original limit expression.✓ Proved
- \[ = \lim_{x \to \infty}\left(\frac{\frac{6}{x} + \frac{1}{x^{2}}}{9 + \frac{3}{x^{2}}}\right) \]algebraDivide the numerator and denominator by x^2.✓ Proved
- \[ = \lim_{x \to \infty}\left(\frac{6}{x} + \frac{1}{x^{2}}\right) \left(\lim_{x \to \infty}\left(9 + \frac{3}{x^{2}}\right)\right)^{-1} \]limit-lawApply the quotient rule for limits.✓ Proved
- \[ = 0 \]limit simplifyEvaluate the limits of the numerator and denominator separately. Simplify the final result.✓ 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
| 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 undefined where 9*x**2 + 3 = 0 undefined where x = 0 undefined where 9 + 3/x**2 = 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where x = 0 undefined where 9 + 3/x**2 = 0 undefined where Limit(9 + 3/x**2, x, oo, dir='-') = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where x = 0 undefined where Limit(9 + 3/x**2, x, oo, dir='-') = 0 |
| 5 | ✓ 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: fail (error) — Step 4 claims to evaluate the limits of the numerator and denominator separately, but it only evaluates the denominator (Limit(9 + 3/x**2) -> 9) while implicitly discarding the numerator's limit (Limit(6/x + 1/x**2) -> 0) without showing the calculation or applying the limit rule to the numerator term. A single step cannot evaluate two distinct sub-expressions into their final values while labeling it as a single 'limit' application on the whole expression, especially when the previous step split them. More critically, the transition from Step 3 to Step 4 skips the actual evaluation of the numerator limit `Limit((6/x + 1/x**2), x, oo)`. The step label 'limit' is insufficient to justify replacing two separate limit expressions with their numerical values in one go without showing the intermediate limit evaluations for each part.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-05 — Step 4 claims to evaluate the limits of the numerator and denominator separately, but it only evaluates the denominator (Limit(9 + 3/x**2) -> 9) while implicitly discarding the numerator's limit (Limit(6/x + 1/x**2) -> 0) without showing the calculation or applying the limit rule to the numerator term. A single step cannot evaluate two distinct sub-expressions into their final values while labeling it as a single 'limit' application on the whole expression, especially when the previous step split them. More critically, the transition from Step 3 to Step 4 skips the actual evaluation of the numerator limit `Limit((6/x + 1/x**2), x, oo)`. The step label 'limit' is insufficient to justify replacing two separate limit expressions with their numerical values in one go without showing the intermediate limit evaluations for each part.gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: pass 2026-10-05gpt-oss:20b: pass 2026-10-05
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 2026-10-05 with SymPy 1.14.0.