Definite Integration
Limit as Definite Integral
Grade 12
Question:
<p>Find \(\displaystyle\lim_{n \to \infty} \left[\frac{1}{n^2}\sec^2\!\left(\frac{1}{n^2}\right) + \frac{2}{n^2}\sec^2\!\left(\frac{4}{n^2}\right) + \cdots + \frac{1}{n^2}\sec^2 1\right]\).</p>
<p>\(\tan 1\)</p>
<p>\(\dfrac{1}{2}\sec^2 1\)</p>
<p>\(\dfrac{1}{2}\tan 1\)</p>
<p>\(\dfrac{1}{2}\sec 1\)</p>
Step-by-Step Solution
Key Concept: Recognize this sum as a Riemann sum for ∫₀¹ sec²(x) dx by setting the k-th term as (1/n²)sec²(k²/n²) with width Δx = 1/n². The substitution u = k/n transforms this into the standard integral form.
<p><strong>Step 1: Rewrite the sum as a Riemann sum</strong></p><p>The given sum is:</p><p>S_n = Σ(k=1 to n) [1/n²·sec²(k²/n²)]</p><p>Reindex: Let r = k/n, so k = rn and Δr = 1/n. Then k²/n² = r².</p><p><strong>Step 2: Recognize the Riemann sum structure</strong></p><p>S_n = Σ(k=1 to n) [1/n·sec²(k²/n²)] where the width is 1/n and we sum over k from 1 to n.</p><p>This is equivalent to: Σ sec²((k/n)²)·(1/n) with partition points k/n ∈ [1/n, 1]</p><p><strong>Step 3: Take the limit as a definite integral</strong></p><p>As n → ∞, this Riemann sum converges to:</p><p>∫₀¹ sec²(x²)·2x dx (after proper substitution u = x²)</p><p>Alternatively, recognizing the direct form: the sum evaluates to ∫₀¹ sec²(x) dx</p><p><strong>Step 4: Evaluate the integral</strong></p><p>∫₀¹ sec²(x) dx = [tan(x)]₀¹ = tan(1) - tan(0) = tan(1)</p><p>∴ Answer: <strong>tan(1)</strong></p>
Correct Answer: C