Limit of \( \displaystyle \frac{e^{4 x + 2} - 1}{\sin{\left(2 x + 1 \right)}} \) as \( x \to - \frac{1}{2} \)
Problem 1.83 · medium
Evaluate \( \displaystyle \lim_{x \to - \frac{1}{2}} \frac{e^{4 x + 2} - 1}{\sin{\left(2 x + 1 \right)}} \).
- \[ \lim_{x \to - \frac{1}{2}^+}\left(\frac{e^{4 x + 2} - 1}{\sin{\left(2 x + 1 \right)}}\right) \]limitIdentify the limit to be evaluated.✓ Proved
- \[ = \lim_{x \to - \frac{1}{2}^+}\left(e^{4 x + 2} - 1\right) \left(\lim_{x \to - \frac{1}{2}^+} \sin{\left(2 x + 1 \right)}\right)^{-1} \]limit-lawCheck if the limit of the numerator and denominator are both non-zero.Not checked
- \[ = \lim_{x \to - \frac{1}{2}^+}\left(\frac{\frac{d}{d x} \left(e^{4 x + 2} - 1\right)}{\frac{d}{d x} \sin{\left(2 x + 1 \right)}}\right) \]lhopitalApply L'Hopital's Rule since the limit is of the form 0/0.✓ Checked independently
- \[ = \lim_{x \to - \frac{1}{2}^+}\left(\frac{2 e^{4 x + 2}}{\cos{\left(2 x + 1 \right)}}\right) \]simplify simplifyCompute the derivatives. Simplify the fraction.✓ Proved
- \[ = 2 \]limit simplify simplify simplifySubstitute the value x = -1/2 into the expression. Simplify the exponents and trigonometric functions. Evaluate the constants. Final result.✓ Proved
Answer \( 2 \)
Lines: 8 proved, 1 checked by independent solution, 1 not checked. 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 | Not checked | claude-sonnet-5-5 | claude-sonnet-5-5: Both limits are 0, so Limit(num)/Limit(den) is 0/0, which is undefined. The quotient limit law needs a non-zero denominator limit, so this step is invalid. It should have been a 'limit' step that identifies the 0/0 form. undefined where sin(2*x + 1) = 0 undefined where Limit(sin(2*x + 1), x, -1/2, dir='+') = 0 |
| 3 | ✓ Checked independently | claude-sonnet-5-5 script, run by sympy 1.14.0 | claude-sonnet-5-5 script, run by sympy 1.14.0: num,den -> 0 (0/0); derivative quotient limit = 2, original limit = 2; L-Hopital hypotheses hold (checks/1.83/line3-fb75e652.py)undefined where Limit(sin(2*x + 1), x, -1/2, dir='+') = 0 undefined where Derivative(sin(2*x + 1), x) = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where Derivative(sin(2*x + 1), x) = 0 undefined where cos(2*x + 1) = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where cos(2*x + 1) = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where cos(2*x + 1) = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 |
| 9 | ✓ 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: inconclusive — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"Step 2 incorrectly applies the limit-law to split the limit of a quotient into the quotient of limits. This rule requires that the denominator’s limit beqwen3.6:27b-mlx: pass
Senior review claude-sonnet-5-5, 2026-10-03: fail — Line 2 applies the quotient limit-law to a 0/0 form, which is invalid. Its note also asks about non-zero limits and then proceeds anyway. The remaining lines (L'Hôpital, then substitution) are correct, and the answer 2 is right. The 0/0 form should have been identified with a limit step, not a limit-law quotient.
gpt-oss:20b: uphold — At x=-1/2 the numerator limit is exp(0)-1=0 and the denominator limit is sin(0)=0, so the quotient limit law does not apply. Line 2 equates the original limit with 0/0, which is false and mislabeled.
Every verdict on record (5)
qwen3.6:27b-mlx: pass 2026-10-03gpt-oss:20b: inconclusive 2026-10-03 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"Step 2 incorrectly applies the limit-law to split the limit of a quotient into the quotient of limits. This rule requires that the denominator’s limit beclaude-sonnet-5-5: fail (error) 2026-10-03 — Line 2 applies the quotient limit-law to a 0/0 form, which is invalid. Its note also asks about non-zero limits and then proceeds anyway. The remaining lines (L'Hôpital, then substitution) are correct, and the answer 2 is right. The 0/0 form should have been identified with a limit step, not a limit-law quotient.qwen3.6:27b-mlx: pass 2026-09-30gpt-oss:20b: fail (error) 2026-09-30 — Step 2 incorrectly applies the limit-law for a quotient when the denominator’s limit is zero. The rule requires the denominator’s limit to be non‑zero, so the subsequent use of L’Hôpital’s rule is unjustified.
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 2026-10-03 with SymPy 1.14.0.