Definite Integration
Limits as Riemann sums
Grade 12
Question:
<p>Find the value of:
\[ L = \lim_{n \to \infty} \sqrt[n]{\frac{n^{2n}}{(n^2+1^2)(n^2+2^2)\cdots(n^2+n^2)}} \]</p>
<p>\(\dfrac{1}{2}e^{2-\pi/2}\)</p>
<p>\(\dfrac{1}{2}e^{\pi/2}\)</p>
<p>\(e^{2-\pi/2}\)</p>
<p>\(\dfrac{1}{2}e^{2+\pi/2}\)</p>
Step-by-Step Solution
Key Concept: Convert the nth root of a product into a sum via logarithms, then recognize the product as a Riemann sum approximation of an integral involving ∏(1 + k²/n²).
<p><strong>Step 1:</strong> Take logarithm: ln L = lim_{n→∞} (1/n)ln[(n^{2n})/∏(n²+k²)] where product is k=1 to n</p><p><strong>Step 2:</strong> Separate: ln L = lim_{n→∞} (1/n)[2n·ln n - ∑ln(n²+k²)] = lim_{n→∞} 2ln n - (1/n)∑ln(n²+k²)</p><p><strong>Step 3:</strong> Rewrite: (1/n)∑ln(n²+k²) = (1/n)∑[2ln n + ln(1+k²/n²)] = 2ln n + (1/n)∑ln(1+k²/n²)</p><p><strong>Step 4:</strong> Recognize Riemann sum: (1/n)∑ln(1+k²/n²) → ∫₀¹ln(1+x²)dx using substitution k/n → x</p><p><strong>Step 5:</strong> Therefore: ln L = 2ln n - 2ln n - ∫₀¹ln(1+x²)dx = -∫₀¹ln(1+x²)dx</p><p><strong>Step 6:</strong> By integration by parts: ∫₀¹ln(1+x²)dx = [xln(1+x²)]₀¹ - ∫₀¹(2x²)/(1+x²)dx = ln 2 - 2[x - arctan x]₀¹ = ln 2 - 2(1 - π/4) = ln 2 - 2 + π/2</p><p><strong>Step 7:</strong> Thus ln L = -(ln 2 - 2 + π/2) = 2 - ln 2 - π/2, so L = e^{2-ln 2-π/2} = (e²/2)·e^{-π/2}</p><p>∴ Answer: C</p>
Correct Answer: C