Definite Integration
Limit as a definite integral (Riemann sum)
Grade 12
Question:
<p>\(\displaystyle\lim_{n\to\infty} \frac{n^2}{\left((n^2+1^2)(n^2+2^2)\cdots(n^2+n^2)\right)^{\frac{1}{n}}}\) equals:</p>
<p>\(2e^{2+\frac{\pi}{2}}\)</p>
<p>\(2e^{2-\frac{\pi}{2}}\)</p>
<p>\(\dfrac{1}{2}e^{2-\frac{\pi}{2}}\)</p>
<p>\(\dfrac{1}{2}e^{2+\frac{\pi}{2}}\)</p>
Step-by-Step Solution
Key Concept: Convert the nth root of a product into a sum using logarithms, then recognize the resulting Riemann sum that converges to a definite integral. Use asymptotic analysis with the integral formula for arctan.
<p><strong>Step 1:</strong> Let L = lim(n→∞) n²/[(n²+1²)(n²+2²)···(n²+n²)]^(1/n). Take logarithm:</p><p>ln L = lim(n→∞) [ln(n²) - (1/n)∑(k=1 to n) ln(n²+k²)]</p><p><strong>Step 2:</strong> Rewrite the sum as a Riemann sum. Factor out n² from each term:</p><p>ln(n²+k²) = ln[n²(1+k²/n²)] = 2ln(n) + ln(1+k²/n²)</p><p><strong>Step 3:</strong> Substitute into the expression:</p><p>ln L = lim(n→∞) [2ln(n) - (1/n)∑(k=1 to n)[2ln(n) + ln(1+k²/n²)]]</p><p>= lim(n→∞) [2ln(n) - 2ln(n) - (1/n)∑(k=1 to n) ln(1+k²/n²)]</p><p><strong>Step 4:</strong> Recognize the Riemann sum (with partition width 1/n, evaluating at k/n):</p><p>ln L = -∫₀¹ ln(1+x²)dx</p><p><strong>Step 5:</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 x·(2x)/(1+x²)dx</p><p>= ln(2) - 2∫₀¹ x²/(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 6:</strong> Therefore:</p><p>ln L = -(ln(2) - 2 + π/2) = 2 - ln(2) - π/2</p><p><strong>Step 7:</strong> Take exponential:</p><p>L = e^(2 - ln(2) - π/2) = e²·e^(-ln(2))·e^(-π/2) = e²/2·e^(-π/2) = (1/2)e^(2-π/2)</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C