Definite Integration
Limit of a sum / improper integrals
Grade 12
Question:
<p><strong>284.</strong> \( L = \lim_{n \to \infty} \sqrt{n} \int_{0}^{1} \dfrac{dx}{(1+x^2)^n} \). Suppose that the above limit exists, then choose the correct option.</p>
<p>(a) \(\dfrac{1}{2} < L < 2\)</p>
<p>(b) \(4 < L < 5\)</p>
<p>(c) \(2 < L \leq 3\)</p>
<p>(d) \(L \geq 5\)</p>
Step-by-Step Solution
Key Concept: The integrand (1+x²)⁻ⁿ approaches 0 rapidly except near x=0 where it equals 1. As n→∞, the integral concentrates near x=0, making substitution u=x√n transform it into a standard form that reveals Laplace's method.
<p><strong>Step 1:</strong> Recognize that (1+x²)⁻ⁿ decays rapidly as n→∞, with maximum value 1 at x=0. Most contribution to the integral comes from x near 0.</p><p><strong>Step 2:</strong> Use substitution u = x√n, so dx = du/√n. The integral becomes:</p><p>∫₀¹ (1+x²)⁻ⁿ dx = (1/√n)∫₀^√n (1+u²/n)⁻ⁿ du</p><p><strong>Step 3:</strong> Therefore:</p><p>L = √n · (1/√n)∫₀^√n (1+u²/n)⁻ⁿ du = ∫₀^√n (1+u²/n)⁻ⁿ du</p><p><strong>Step 4:</strong> As n→∞, (1+u²/n)⁻ⁿ → e⁻ᵘ² and the upper limit √n→∞:</p><p>L = ∫₀^∞ e⁻ᵘ² du = √π/2</p><p>∴ Answer: A</p>
Correct Answer: A