Definite Integration
Limit as Definite Integral
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: Recognize the product under the nth root as a Riemann sum approximation. Convert the logarithm of the geometric mean into an integral: (1/n)∑ln(n²+k²) → ∫ln(1+x²)dx by substituting k=nx.
<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→∞) [2ln n - (1/n)∑(k=1 to n) ln(n²+k²)]</p><p><strong>Step 2:</strong> Rewrite the product sum: (1/n)∑ln(n²+k²) = (1/n)∑ln(n²[1+(k/n)²]) = (1/n)∑[2ln n + ln(1+(k/n)²)]</p><p>= 2ln n + (1/n)∑ln(1+(k/n)²)</p><p><strong>Step 3:</strong> Recognize (1/n)∑ln(1+(k/n)²) as a Riemann sum with Δx=1/n, xₖ=k/n:</p><p>lim(n→∞) (1/n)∑(k=1 to n) ln(1+(k/n)²) = ∫₀¹ ln(1+x²)dx</p><p><strong>Step 4:</strong> Evaluate ∫₀¹ ln(1+x²)dx using integration by parts: u=ln(1+x²), dv=dx</p><p>= [x·ln(1+x²)]₀¹ - ∫₀¹ x·(2x)/(1+x²)dx = ln 2 - 2∫₀¹ x²/(1+x²)dx</p><p>= ln 2 - 2∫₀¹ [1 - 1/(1+x²)]dx = ln 2 - 2[x - arctan x]₀¹</p><p>= ln 2 - 2(1 - π/4) = ln 2 - 2 + π/2</p><p><strong>Step 5:</strong> Therefore: ln L = 2ln n - 2ln n - (ln 2 - 2 + π/2) = 2 - π/2 - ln 2</p><p>∴ L = e^(2-π/2-ln 2) = e²/(2e^(π/2))</p><p><strong>Answer: C</strong></p>
Correct Answer: C