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Question #41504

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the limit: L=lim⁡n→∞nπ∫−11(1−x2)n1+eπxcos⁡(x)dxL = \lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{\pi}} \int_{-1}^{1} \frac{(1-x^2)^n}{1+e^{\pi x}} \cos(x) dx

Options

  • A.

    12\frac{1}{2}

  • B.

    1

  • C.

    1π\frac{1}{\pi}

  • D.

    π2\frac{\pi}{2}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Trigonometric Identical Substitutions|Symmetry Utilization|Squeeze Theorem (Sandwich Theorem)

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of symmetry properties in definite integrals, trigonometric substitution, and the Squeeze Theorem for asymptotic evaluation.

The primary trap lies in the resulting integral ∫0π/2cos⁡2n+1θcos⁡(sin⁡θ)dθ\int_{0}^{\pi/2} \cos^{2n+1} \theta \cos(\sin \theta) d\theta, which has no elementary antiderivative.

Students relying purely on procedural integration techniques will attempt integration by parts or series expansion, leading to a time-consuming dead end.

The correct architectural pivot requires recognizing that the integrand can be bounded using 1−12sin⁡2θ≤cos⁡(sin⁡θ)≤11 - \frac{1}{2}\sin^2 \theta \le \cos(\sin \theta) \le 1, allowing the Squeeze Theorem to trap the integral between two evaluable Wallis' integrals.

This punishes procedural reliance and rewards conceptual bounding and asymptotic analysis.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterIntegrals - Class 12
ConceptTrigonometric Identical Substitutions
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree