<p>Find the following limit: <br> 16. \(\lim_{n \to \infty} \left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right)\left(1+\frac{3^2}{n^2}\right)\cdots\left(1+\frac{n^2}{n^2}\right)\right]^{1/n}\)</p>
Step-by-Step Solution
Key Concept: Convert the nth root of a product into a Riemann sum by taking logarithm: ln(product)^(1/n) = (1/n)∑ln(1+k²/n²). Recognize this as a Riemann sum for ∫₀¹ ln(1+x²)dx, then exponentiate the result.
<p><strong>Step 1:</strong> Let L = lim(n→∞) [(1+1/n²)(1+4/n²)⋯(1+n²/n²)]^(1/n). Take natural logarithm:</p><p>ln(L) = lim(n→∞) (1/n)∑_{k=1}^{n} ln(1+k²/n²)</p><p><strong>Step 2:</strong> Recognize this as a Riemann sum with partition width Δx = 1/n and sample points x_k = k/n:</p><p>ln(L) = lim(n→∞) (1/n)∑_{k=1}^{n} ln(1+(k/n)²) = ∫₀¹ ln(1+x²)dx</p><p><strong>Step 3:</strong> Evaluate ∫₀¹ ln(1+x²)dx using integration by parts with u = ln(1+x²), dv = dx:</p><p>∫₀¹ ln(1+x²)dx = [x·ln(1+x²)]₀¹ - ∫₀¹ 2x²/(1+x²)dx</p><p>= ln(2) - 2∫₀¹ [1 - 1/(1+x²)]dx</p><p>= ln(2) - 2[x - arctan(x)]₀¹</p><p>= ln(2) - 2(1 - π/4) = ln(2) - 2 + π/2</p><p><strong>Step 4:</strong> Exponentiate to find L:</p><p>L = e^(ln(2) - 2 + π/2) = 2·e^(π/2 - 2)</p><p>∴ Answer: <strong>2e^(π/2 - 2)</strong> or equivalently <strong>e^(ln(2) + π/2 - 2)</strong></p>
Correct Answer: 2