<p>Find the following limit: <br> 15. \(\lim_{n \to \infty} \left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\left(1+\frac{3}{n}\right)\cdots\left(1+\frac{n}{n}\right)\right]^{1/n}\)</p>
Step-by-Step Solution
Key Concept: Convert the product into a sum using logarithms, then recognize the resulting limit as a Riemann sum that equals a definite integral of ln(1+x) from 0 to 1.
<p><strong>Step 1:</strong> Let L = lim(n→∞) [(1+1/n)(1+2/n)···(1+n/n)]^(1/n). Take natural logarithm:</p><p>ln L = lim(n→∞) (1/n) Σₖ₌₁ⁿ ln(1+k/n)</p><p><strong>Step 2:</strong> Recognize this as a Riemann sum with Δx = 1/n and partition points k/n. As n→∞:</p><p>ln L = ∫₀¹ ln(1+x)dx</p><p><strong>Step 3:</strong> Evaluate using integration by parts. Let u = ln(1+x), dv = dx:</p><p>∫₀¹ ln(1+x)dx = [x·ln(1+x)]₀¹ - ∫₀¹ x/(1+x)dx</p><p>= ln(2) - ∫₀¹ [1 - 1/(1+x)]dx</p><p>= ln(2) - [x - ln(1+x)]₀¹</p><p>= ln(2) - [1 - ln(2)]</p><p>= 2ln(2) - 1 = ln(4/e)</p><p><strong>Step 4:</strong> Therefore ln L = ln(4/e), so L = 4/e</p><p>∴ Answer: <strong>4/e</strong></p>
Correct Answer: 4