Limit of \( \displaystyle \frac{\ln{\left(3 x - 1 \right)}}{3 x - 1} \) as \( x \to \infty \)
Problem 1.375 · medium
Evaluate \( \displaystyle \lim_{x \to \infty} \frac{\ln{\left(3 x - 1 \right)}}{3 x - 1} \).
- \[ \lim_{x \to \infty}\left(\frac{\ln{\left(3 x - 1 \right)}}{3 x - 1}\right) \]limitStart with the limit of the given function.✓ Proved
- \[ = \lim_{x \to \infty}\left(\frac{\ln{\left(3 x \left(1 - \frac{1}{3 x}\right) \right)}}{3 x \left(1 - \frac{1}{3 x}\right)}\right) \]rewriteRewrite the expression to isolate the dominant term as x approaches infinity.✓ Proved
- \[ = \lim_{x \to \infty}\left(\frac{\ln{\left(3 x \right)} + \ln{\left(1 - \frac{1}{3 x} \right)}}{3 x \left(1 - \frac{1}{3 x}\right)}\right) \]algebraUse the logarithm product rule.✓ Proved
- \[ = \lim_{x \to \infty}\left(\frac{\ln{\left(3 x \right)}}{3 x \left(1 - \frac{1}{3 x}\right)} + \frac{\ln{\left(1 - \frac{1}{3 x} \right)}}{3 x \left(1 - \frac{1}{3 x}\right)}\right) \]algebraSplit the fraction into two parts.✓ Proved
- \[ = \lim_{x \to \infty}\left(\frac{\ln{\left(3 x \right)}}{3 x \left(1 - \frac{1}{3 x}\right)}\right) + \lim_{x \to \infty}\left(\frac{\ln{\left(1 - \frac{1}{3 x} \right)}}{3 x \left(1 - \frac{1}{3 x}\right)}\right) \]limit-lawApply the sum rule for limits.✓ Proved
- \[ = \lim_{x \to \infty} 0 + \lim_{x \to \infty}\left(\frac{\ln{\left(3 x \right)}}{3 x}\right) \]limit-lawEvaluate the limits of the individual terms separately.✓ Proved
- \[ = 0 \]limit limitThe first part goes to 0 by growth rates, and the second part is 0 because log(1)=0. 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. 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 log is undefined for non-positive arguments undefined where 3*x - 1 = 0 undefined where x = 0 undefined where 1 - 1/(3*x) = 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x = 0 undefined where 1 - 1/(3*x) = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x = 0 undefined where 1 - 1/(3*x) = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x = 0 undefined where 1 - 1/(3*x) = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x = 0 undefined where 1 - 1/(3*x) = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x = 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 6 incorrectly simplifies the second term: it replaces ∑…/(3x(1-1/(3x))) with ∑…/(3x) and then with ∑…/(1), ignoring the denominator’s limit. The correct limit of that term is 0 because the numerator tends to 0 and the denominator tends to infinity, not simply ∑…/(1).qwen3.6:27b-mlx: fail (error) — Step 6 incorrectly simplifies the second term's limit to Limit(log(1), x, oo). The term is log(1 - 1/(3*x)) / (3*x*(1 - 1/(3*x))). As x -> oo, the numerator approaches log(1)=0 and the denominator approaches infinity, so the limit is 0. However, the step replaces the entire fraction with just log(1), ignoring the denominator, which is mathematically invalid (it changes the expression from 0/infinity to 0). While the final value 0 is correct, the intermediate expression is wrong.
Senior review claude-sonnet-5-5, 2026-10-07: fail — Step 6 equates Limit(log(1-1/(3x))/(3x(1-1/(3x)))) with Limit(log(1)), which are different functions that only happen to share the limit 0 (0/infinity form). It also silently drops the factor (1-1/(3x)) from the first term. This teaches that you can replace pieces of a quotient by their limits and drop a divergent denominator. A valid route would factor out 1/(1-1/(3x)) as a product limit, or bound the second term, in separate steps.
gpt-oss:20b: uphold — Step 6 replaces the whole second term with Limit(log(1)), dropping the denominator 3x(1-1/(3x)), and also removes the (1-1/(3x)) factor from the first term; these are several independent changes, not one limit-law application.qwen3.6:27b-mlx: uphold — Its core point is correct: step 6 substitutes limits of sub-pieces inside the Limit operator, which is not the sum/product/quotient limit law and changes two terms. Its Step 3 domain worry is dismissed, since domain restrictions are never a defect.
Every verdict on record (5)
gpt-oss:20b: fail (error) 2026-10-07 — Step 6 incorrectly simplifies the second term: it replaces ∑…/(3x(1-1/(3x))) with ∑…/(3x) and then with ∑…/(1), ignoring the denominator’s limit. The correct limit of that term is 0 because the numerator tends to 0 and the denominator tends to infinity, not simply ∑…/(1).qwen3.6:27b-mlx: fail (error) 2026-10-07 — Step 6 incorrectly simplifies the second term's limit to Limit(log(1), x, oo). The term is log(1 - 1/(3*x)) / (3*x*(1 - 1/(3*x))). As x -> oo, the numerator approaches log(1)=0 and the denominator approaches infinity, so the limit is 0. However, the step replaces the entire fraction with just log(1), ignoring the denominator, which is mathematically invalid (it changes the expression from 0/infinity to 0). While the final value 0 is correct, the intermediate expression is wrong.claude-sonnet-5-5: fail (misleading) 2026-10-07 — Step 6 equates Limit(log(1-1/(3x))/(3x(1-1/(3x)))) with Limit(log(1)), which are different functions that only happen to share the limit 0 (0/infinity form). It also silently drops the factor (1-1/(3x)) from the first term. This teaches that you can replace pieces of a quotient by their limits and drop a divergent denominator. A valid route would factor out 1/(1-1/(3x)) as a product limit, or bound the second term, in separate steps.qwen3.6:27b-mlx: fail (style) 2026-10-07 — [domain objection, downgraded to style] Step 6 incorrectly simplifies the second term to Limit(log(1), x, oo) without justification, effectively assuming the limit of the denominator is 1 and the numerator is 0 simultaneously, which is not a valid single-step application of limit laws. Additionally, Step 3 applies the product rule for logarithms, but the argument 3*x*(1 - 1/(3*x)) is not strictly positive for all x in the domain approaching infinity if we consider complex branches or specific real intervals, though more importantly, the transition from Step 2 to 3 assumes log(ab) = log(a) + log(b) which is an algebraic identity, but Step 6 is the primary error where the limit of a quotient is replaced by the quotient of limits without verifying the denominator's limit is non-zero (it is 1, so it's technically valid, but the notation 'Limit(log(1), x, oo)' implies the term became constant, which is a leap from the previous line). Actually, the bigger issue is Step 6: The term log(1 - 1/(3*x))/(3*x*(1 - 1/(3*x))) does not become log(1). The numerator approaches log(1)=0 and the denominator approaches infinity. The step claims the limit is Limit(log(1), x, oo) which is 0. This is a 'simplify' or 'substitution' step disguised as a limit law, but it skips the evaluation of the denominator. More critically, Step 6 applies 'limit-law' to split the limit, but the note says 'Evaluate the limits... separately'. The transformation from Step 5 to 6 is not just applying a limit law; it is evaluating the inner limits. Specifically, the second term in Step 5 is Limit(log(1 - 1/(3*x))/(3*x*(1 - 1/(3*x))), x, oo). Step 6 replaces this with Limit(log(1), x, oo). This is mathematically incorrect as a direct substitution because the denominator is not 1, it is approaching 1, but the whole fraction approaches 0/infinity = 0. Replacing the whole expression with log(1) is wrong. The limit of the second term is 0, but it is not equal to Limit(log(1), x, oo) in the sense of algebraic equality of the expressions inside the limit operator before evaluation. The expression inside the limit in step 6 is constant 0, whereas in step 5 it is a function of x. This is a 'rewrite' or 'simplify' step, not a 'limit-law' step. Furthermore, Step 6 claims the first term becomes Limit(log(3*x)/(3*x), x, oo). The original first term in Step 5 was Limit(log(3*x)/(3*x*(1 - 1/(3*x))), x, oo). The factor (1 - 1/(3*x)) in the denominator approaches 1. So the limit is the same. But the step label 'limit-law' usually refers to sum/product/quotient rules for limits, not simplifying the integrand. The step combines simplification of the expression inside the limit with the application of limit properties. This violates the 'one thing per step' rule. It simplifies the expression AND applies limit properties (or prepares for them). Specifically, it changes the expression inside the limit. That is 'algebra' or 'rewrite'. It is not 'limit-law'. 'Limit-law' applies to the Limit operator itself (e.g., Limit(f+g) = Limit(f) + Limit(g)). Step 5 did that. Step 6 changes the functions inside. That is a defect in labeling and granularity.gpt-oss:20b: fail (error) 2026-10-07 — Step 6 incorrectly replaces the fraction log(1-1/(3*x))/(3*x*(1-1/(3*x))) with log(1) and drops the denominator, applying two simplifications at once and violating the rule that each step must change only one thing.
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 2026-10-07 with SymPy 1.14.0.