Limit of \( \displaystyle \frac{\ln{\left(2 x \right)}}{2 x - 1} \) as \( x \to \frac{1}{2} \)
Problem 1.636 · medium Video
Evaluate \( \displaystyle \lim_{x \to \frac{1}{2}} \frac{\ln{\left(2 x \right)}}{2 x - 1} \).
- \[ \lim_{x \to \frac{1}{2}^+}\left(\frac{\ln{\left(2 x \right)}}{2 x - 1}\right) \]limitIdentify the limit type.✓ Proved
- \[ = \lim_{x \to \frac{1}{2}^+}\left(\frac{\frac{d}{d x} \ln{\left(2 x \right)}}{\frac{d}{d x} \left(2 x - 1\right)}\right) \]lhopitalApply L'Hopital's rule since the limit is an indeterminate form 0/0.✓ Proved
- \[ = \lim_{x \to \frac{1}{2}^+}\left(\frac{1}{2 x}\right) \]simplify simplifyCompute the derivatives. Simplify the expression.✓ Proved
- \[ = 1 \]limit simplifyEvaluate the limit by substituting x = 1/2. Final simplification.✓ Proved
Answer \( 1 \)
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. How we verify
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 2*x - 1 = 0 undefined where Derivative(2*x - 1, x) = 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where Derivative(2*x - 1, x) = 0 undefined where x = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where x = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where x = 0 |
| 6 | ✓ 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 2 incorrectly applies L’Hôpital’s rule: the derivative of log(2x) is 1/x (not 1/(2x)), and the derivative of 2x‑1 is 2 (not 1). This step applies two incorrect rules at once and gives an incorrect intermediate expression.qwen3.6:27b-mlx: pass
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17gpt-oss:20b: fail (error) 2026-10-17 — Step 2 incorrectly applies L’Hôpital’s rule: the derivative of log(2x) is 1/x (not 1/(2x)), and the derivative of 2x‑1 is 2 (not 1). This step applies two incorrect rules at once and gives an incorrect intermediate expression.qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the 0/0 indeterminate form and applies L'Hôpital's rule. The subsequent differentiation and simplification steps are algebraically correct and properly labeled.gpt-oss:20b: fail (error) 2026-10-15 — Step 2 applies L’Hôpital’s rule but incorrectly differentiates the numerator. The derivative of log(2*x) is 1/x, not 1/(2*x).qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly identifies the indeterminate form and applies L'Hopital's Rule. Each step changes only one aspect of the expression (applying the rule, computing derivatives, simplifying, substituting), and the labels used are from the allowed vocabulary and accurately describe the operations performed.gpt-oss:20b: fail (error) 2026-10-14 — Step 3 incorrectly applies the derivative of log(2*x); it should be 1/x, not 1/(2*x).
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 2026-10-17 with SymPy 1.14.0.