Limit of \( \displaystyle \frac{1 - \cos{\left(2 x - 1 \right)}}{\left(2 x - 1\right)^{2}} \) as \( x \to \frac{1}{2} \)
Problem 1.498 · medium
Evaluate \( \displaystyle \lim_{x \to \frac{1}{2}} \frac{1 - \cos{\left(2 x - 1 \right)}}{\left(2 x - 1\right)^{2}} \).
- \[ \lim_{x \to \frac{1}{2}^+}\left(\frac{1 - \cos{\left(2 x - 1 \right)}}{\left(2 x - 1\right)^{2}}\right) \]limit algebra algebraStart with the original limit. Rewrite the argument of the cosine. Factor out 2 from the denominator.✓ Proved
- \[ = \lim_{x \to \frac{1}{2}^+}\left(\frac{\frac{d}{d x} \left(1 - \cos{\left(2 x - 1 \right)}\right)}{\frac{d}{d x} \left(2 x - 1\right)^{2}}\right) \]lhopitalApply L'Hopital's rule.✓ Proved
- \[ = \lim_{x \to \frac{1}{2}^+}\left(\frac{2 \sin{\left(2 x - 1 \right)}}{8 x - 4}\right) \]simplify simplifyCompute the derivatives. Simplify the denominator.✓ Proved
- \[ = \lim_{x \to \frac{1}{2}^+}\left(\frac{\sin{\left(2 x - 1 \right)}}{4 x - 2}\right) \]simplifyCancel the common factor of 2.✓ Proved
- \[ = \lim_{u \to 0^+}\left(\frac{\sin{\left(2 u \right)}}{4 u}\right) \]substitution algebra simplifyLet u = x - 1/2, so x = u + 1/2. Substitute x = u + 1/2 into the expression. Simplify the denominator.✓ Proved
- \[ = \lim_{u \to 0^+} \frac{1}{2} \]trig-limitUse the fundamental trigonometric limit sin(z)/z -> 1 as z -> 0.✓ Proved
- \[ = \frac{1}{2} \]limitEvaluate the limit.✓ Proved
Answer \( \frac{1}{2} \)
✓ Nihil obstat 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. Reviewers found nothing wrong with the explanation.
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 2*x - 1 = 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where 2*x - 1 = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where 2*x - 1 = 0 undefined where Derivative((2*x - 1)**2, x) = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where Derivative((2*x - 1)**2, x) = 0 undefined where 8*x - 4 = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where 8*x - 4 = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where 8*x - 4 = 0 undefined where 4*x - 2 = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where 4*x - 2 = 0 undefined where u = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where u = 0 |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where u = 0 |
| 11 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where u = 0 |
| 12 | ✓ 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: pass — The solution correctly applies L'Hopital's rule and algebraic simplifications to evaluate the limit. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.
Senior review claude-sonnet-5-5, 2026-10-10: pass — Every line was proved equivalent by SymPy, and the labels name rules the steps legitimately apply. Minor style points remain: line 8 combines the substitution with simplifying the sine's argument, the note on line 9 describes line 8's work, and line 11 would read better as an explicit factoring of (1/2)·sin(2u)/(2u) before applying the trig limit.
qwen3.6:27b-mlx: dismiss — The suggested label `derivative` is not in the allowed vocabulary, so `simplify` for evaluating the derivatives written in line 4 is acceptable. The derivatives themselves are correct (2 sin(2x-1) and 2·2·(2x-1)), and SymPy proved the line, so the objection identifies no real defect.
Every verdict on record (5)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly applies L'Hopital's rule and algebraic simplifications to evaluate the limit. Each step changes only one aspect of the expression and uses valid labels from the fixed vocabulary.claude-sonnet-5-5: pass 2026-10-10 — Every line was proved equivalent by SymPy, and the labels name rules the steps legitimately apply. Minor style points remain: line 8 combines the substitution with simplifying the sine's argument, the note on line 9 describes line 8's work, and line 11 would read better as an explicit factoring of (1/2)·sin(2u)/(2u) before applying the trig limit.qwen3.6:27b-mlx: fail (error) 2026-10-10 — Step 4 applies L'Hôpital's rule to the expression from Step 3, but Step 3 contains a rewritten numerator `1 - cos(2*(x - 1/2))` that was not present in the original limit or Step 2's denominator context for differentiation. More critically, Step 4 differentiates the expression from Step 3, but the numerator in Step 3 is `1 - cos(2*(x - 1/2))`. The derivative of this with respect to x is `2*sin(2*(x - 1/2))`. However, Step 5 shows the numerator as `2*sin(2*x - 1)`, which is algebraically equivalent, but the step label `simplify` is used for computing derivatives, which is incorrect; `derivative` should be used. Furthermore, Step 4 jumps from the algebraic manipulation in Step 3 to differentiation, but the expression in Step 3 is not in the form required for direct L'Hôpital application without first confirming the 0/0 form, which is implicit but the transition from Step 3 to 4 is abrupt. The primary error is that Step 4 applies L'Hôpital's rule to the expression in Step 3, but the note says 'Apply L'Hopital's rule' while the step itself shows the result of differentiation. The label `lhopital` is correct for the rule applied, but the step combines the decision to apply L'Hôpital with the computation of derivatives? No, Step 4 shows the limit of the ratio of derivatives. Step 5 computes them. This is acceptable granularity. Let's look closer at Step 4. It transforms `Limit((1 - cos(2*(x - 1/2)))/(2*(x - 1/2))**2, x, 1/2)` into `Limit(Derivative(...) / Derivative(...), x, 1/2)`. This is a valid application of L'Hôpital's rule. Let's look at Step 5. It computes the derivatives. The numerator derivative is `d/dx (1 - cos(2x - 1)) = 2 sin(2x - 1)`. The denominator derivative is `d/dx (2x - 1)^2 = 2(2x - 1)*2 = 4(2x - 1)`. Step 5 shows `(2*sin(2*x - 1)) / (2 * 2 * (2*x - 1))`. This is correct. The label is `simplify`. Computing a derivative is not simplification. It is differentiation. The label should be `derivative`. Let's look at Step 11. It replaces `sin(2*u)` with `2*u`. This is an application of the limit `sin(z)/z -> 1`. The step goes from `Limit(sin(2*u) / (4*u), u, 0)` to `Limit((2*u) / (4*u), u, 0)`. This is effectively using `sin(2u) ~ 2u`. The label `trig-limit` is appropriate. The main defect is Step 5's label. It performs differentiation, not simplification. The vocabulary includes `derivative`. Using `simplify` for differentiation is a label error.gpt-oss:20b: pass 2026-10-10
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.