<p>Let \(f(x) = \lim_{n \to \infty} \dfrac{\cos x}{1 + (\tan^{-1} x)^n}\), then the value of \(\displaystyle\int_0^{\infty} f(x)\, dx\) is equal to:</p>
<p>(a) \(\cos(\tan 1)\)</p>
<p>(b) \(\sin(\tan 1)\)</p>
<p>(c) \(\tan(\tan 1)\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: The limit function f(x) behaves differently depending on whether |tan⁻¹(x)| is less than, equal to, or greater than 1. Since tan⁻¹(x) ranges from 0 to π/2 on [0,∞), the function has a discontinuity at x=1 where tan⁻¹(1)=π/4, creating a piecewise function that must be integrated in two parts.
<p><strong>Step 1: Analyze the limit behavior</strong></p><p>For the limit of (tan⁻¹x)ⁿ as n→∞:</p><ul><li>When 0 ≤ tan⁻¹(x) < 1 (i.e., 0 ≤ x < 1): (tan⁻¹x)ⁿ → 0, so f(x) = cos(x)/1 = cos(x)</li><li>When tan⁻¹(x) = 1 (i.e., x = 1): (tan⁻¹x)ⁿ = 1, so f(1) = cos(1)/2</li><li>When tan⁻¹(x) > 1 (i.e., x > 1): (tan⁻¹x)ⁿ → ∞, so f(x) = cos(x)/∞ = 0</li></ul><p><strong>Step 2: Set up the integral</strong></p><p>∫₀^∞ f(x)dx = ∫₀¹ cos(x)dx + ∫₁^∞ 0·dx</p><p><strong>Step 3: Evaluate</strong></p><p>∫₀¹ cos(x)dx = [sin(x)]₀¹ = sin(1) - sin(0) = sin(1)</p><p>The point x=1 (a single point) has measure zero and doesn't affect the integral value.</p><p>∴ Answer: <strong>sin(1)</strong> (Option B)</p>
Correct Answer: B