Definite Integration
Limits involving geometric mean and definite integrals
Grade 12
Question:
<p>For \(n \geq 1\), Let \(G_n\) be the geometric mean of \(\left\{\sin\frac{k\pi}{2n} : 1 \leq k \leq n\right\}\), then \(\lim_{n \to \infty} G_n\) equals:</p><p>[Note: [k] denotes greatest integer function less than or equal to k.]</p>
<p>\(\lim_{x \to 0} \dfrac{1 - \cos x}{x}\)</p>
<p>\(\lim_{x \to 0} \dfrac{1 - \cos x}{x^2}\)</p>
<p>\(\dfrac{2}{\pi} \displaystyle\int_0^{\pi/2} \sin^2 x\, dx\)</p>
<p>\(\lim_{x \to 0^-} \left[\dfrac{e^x - 1}{x}\right]\)</p>
Step-by-Step Solution
Key Concept: The geometric mean of n terms equals the nth root of their product. For sin values, taking logarithms converts the product into a sum, which can be approximated by a Riemann sum that evaluates to a definite integral as n→∞.
<p><strong>Step 1:</strong> Express the geometric mean as: G_n = [∏(k=1 to n) sin(kπ/2n)]^(1/n)</p><p><strong>Step 2:</strong> Take logarithm: ln(G_n) = (1/n)∑(k=1 to n) ln(sin(kπ/2n))</p><p><strong>Step 3:</strong> Recognize this as a Riemann sum. Let f(x) = ln(sin(x)) with partition points at kπ/(2n). As n→∞, this approaches the definite integral:</p><p>lim(n→∞) ln(G_n) = ∫₀^(π/2) ln(sin(x)) dx · (2/π)</p><p><strong>Step 4:</strong> The integral ∫₀^(π/2) ln(sin(x)) dx = -(π/2)ln(2) (standard result)</p><p><strong>Step 5:</strong> Therefore: lim(n→∞) ln(G_n) = (2/π) · (-(π/2)ln(2)) = -ln(2)</p><p><strong>Step 6:</strong> Taking exponential: lim(n→∞) G_n = e^(-ln(2)) = 1/2</p><p>∴ Answer: C (which equals 1/2)</p>
Correct Answer: C