Definite Integration
Limits as Riemann sums / Geometric mean
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>(a) \(\lim_{x\to 0}\dfrac{1-\cos x}{x}\)</p>
<p>(b) \(\lim_{x\to 0}\dfrac{1-\cos x}{x^2}\)</p>
<p>(c) \(\dfrac{2}{\pi}\displaystyle\int_0^{\pi/2} \sin^2 x\, dx\)</p>
<p>(d) \(\lim_{x\to 0^-}\left[\dfrac{e^x - 1}{x}\right]\)</p>
Step-by-Step Solution
Key Concept: Convert the geometric mean into a product, take logarithm to convert it to a sum, recognize this as a Riemann sum for the integral of ln(sin x), and evaluate using the known result ∫₀^(π/2) ln(sin x)dx = -π ln(2)/2.
<p><strong>Step 1:</strong> Write the geometric mean as a product.</p><p>$$G_n = \left(\prod_{k=1}^{n} \sin\frac{k\pi}{2n}\right)^{1/n}$$</p><p><strong>Step 2:</strong> Take natural logarithm and convert to a sum.</p><p>$$\ln G_n = \frac{1}{n}\sum_{k=1}^{n} \ln\left(\sin\frac{k\pi}{2n}\right)$$</p><p><strong>Step 3:</strong> Recognize this as a Riemann sum. Let $x_k = \frac{k\pi}{2n}$, so $\Delta x = \frac{\pi}{2n}$.</p><p>$$\ln G_n = \frac{1}{\pi} \cdot \frac{\pi}{2} \sum_{k=1}^{n} \ln\left(\sin x_k\right) \cdot \frac{\pi}{2n}$$</p><p>As $n \to \infty$, this becomes:</p><p>$$\lim_{n\to\infty} \ln G_n = \frac{1}{\pi} \int_0^{\pi/2} \ln(\sin x)\,dx \cdot \frac{\pi}{2} = \frac{1}{2}\int_0^{\pi/2} \ln(\sin x)\,dx$$</p><p><strong>Step 4:</strong> Evaluate the standard integral: $\int_0^{\pi/2} \ln(\sin x)\,dx = -\frac{\pi\ln 2}{2}$</p><p>$$\lim_{n\to\infty} \ln G_n = \frac{1}{2} \cdot \left(-\frac{\pi\ln 2}{2}\right) = -\frac{\pi\ln 2}{4}$$</p><p><strong>Step 5:</strong> Exponentiate to find $G_n$.</p><p>$$\lim_{n\to\infty} G_n = e^{-\pi\ln 2/4} = 2^{-\pi/4}$$</p><p>∴ Answer: C</p>
Correct Answer: C