Definite Integration
Limit as sum / Limiting value of integral
Grade 12
Question:
<p>Find a real number \(c\) and a positive number \(L\) for which \(\displaystyle\lim_{r \to \infty} \dfrac{r^c \int_0^{\pi/2} x^r \sin x\,dx}{\int_0^{\pi/2} x^r \cos x\,dx} = L\)</p>
<p>\(c = 1,\, L = \dfrac{2}{\pi}\)</p>
<p>\(c = -1,\, L = \dfrac{2}{\pi}\)</p>
<p>\(c = -1,\, L = \dfrac{4}{\pi}\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: As r→∞, the integrals are dominated by the behavior near x=π/2 where x is maximum. Use Laplace's method or substitute u=π/2-x to analyze the asymptotic behavior, then balance the powers of r in numerator and denominator.
<p><strong>Step 1: Identify the asymptotic regime</strong></p><p>For large r, both ∫₀^(π/2) xʳ sin x dx and ∫₀^(π/2) xʳ cos x dx are dominated by values near x = π/2 where x is largest.</p><p><strong>Step 2: Apply Laplace's method</strong></p><p>Substitute u = π/2 - x, so x = π/2 - u. As r→∞:</p><p>∫₀^(π/2) xʳ sin x dx ≈ ∫₀^(π/2) (π/2 - u)ʳ cos u du ≈ (π/2)ʳ ∫₀^(π/2) e^(r ln(1-2u/π)) cos u du</p><p>For small u: (π/2 - u)ʳ ≈ (π/2)ʳ e^(-2ru/π)</p><p>∴ ∫₀^(π/2) xʳ sin x dx ~ (π/2)ʳ · (π/2r) · 1 ~ (π²/4r)(π/2)ʳ</p><p><strong>Step 3: Similarly for the denominator</strong></p><p>∫₀^(π/2) xʳ cos x dx ~ (π/2)ʳ · (π/2r) · 1 ~ (π²/4r)(π/2)ʳ</p><p><strong>Step 4: More careful asymptotic analysis</strong></p><p>Using integration by parts or Laplace's method more precisely:</p><p>∫₀^(π/2) xʳ sin x dx ~ (π/2)ʳ⁺¹/(r+1)</p><p>∫₀^(π/2) xʳ cos x dx ~ (π/2)ʳ/(r+1)</p><p><strong>Step 5: Form the ratio</strong></p><p>rᶜ · [(π/2)ʳ⁺¹/(r+1)] / [(π/2)ʳ/(r+1)] = rᶜ · (π/2)</p><p>For this limit to be finite and nonzero, we need: <strong>c = 0</strong> and <strong>L = π/2</strong></p><p>∴ Answer: C (c=0, L=π/2)
Correct Answer: C