Definite Integration
Limit as sum / limiting value of integrals
Grade 12
Question:
<p>\( L = \lim_{n \to \infty} \sqrt{n} \int_{0}^{1} \dfrac{dx}{(1+x^2)^n} \). 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 integral concentrates near x=0 where (1+x²)^n is largest; use substitution x=t/√n to convert the limit into a Riemann integral form, revealing that the integral behaves like ∫₀^∞ e^(-t²) dt as n→∞.
<p><strong>Step 1:</strong> Recognize that as n→∞, (1+x²)^n decays rapidly except near x=0. The integral is dominated by the behavior near x=0.</p><p><strong>Step 2:</strong> Substitute x = t/√n, so dx = dt/√n. Then:</p><p>L = lim_{n→∞} √n · ∫₀^(√n) 1/(1+t²/n)^n · (dt/√n) = lim_{n→∞} ∫₀^(√n) 1/(1+t²/n)^n dt</p><p><strong>Step 3:</strong> As n→∞, (1+t²/n)^n → e^(t²) for fixed t. By dominated convergence:</p><p>L = ∫₀^∞ e^(-t²) dt = √π/2</p><p><strong>Step 4:</strong> Therefore L = √π/2 ≈ 0.886.</p><p>∴ Answer: A</p>
Correct Answer: A