<p>Let \( f(x) = \lim_{n \to \infty} (\sin x)^{2n} \), then \( f \) is</p>
<p>(a) continuous at \( x = \pi/2 \)</p>
<p>(b) discontinuous \( x = \pi/2 \)</p>
<p>(c) discontinuous at \( x = -\pi/2 \)</p>
<p>(d) discontinuous at an infinite number of points</p>
Step-by-Step Solution
Key Concept: Recognize that (sin x)^(2n) converges to 0 when |sin x| < 1 and to 1 when |sin x| = 1 as n→∞. This creates a piecewise function with jump discontinuities at odd multiples of π/2.
<p><strong>Step 1:</strong> Analyze the limit based on the value of sin x.</p><p>For any fixed x, consider (sin x)^(2n) as n → ∞:</p><ul><li>If |sin x| < 1: (sin x)^(2n) → 0</li><li>If |sin x| = 1 (i.e., sin x = ±1): (sin x)^(2n) → 1</li><li>If sin x = 0: (sin x)^(2n) = 0</li></ul><p><strong>Step 2:</strong> Determine where |sin x| = 1.</p><p>sin x = 1 when x = (4k+1)π/2 (i.e., x = π/2, 5π/2, ...)</p><p>sin x = -1 when x = (4k+3)π/2 (i.e., x = 3π/2, 7π/2, ...)</p><p>Combined: x = (2k+1)π/2 for k ∈ ℤ</p><p><strong>Step 3:</strong> Write the piecewise function.</p><p>f(x) = {1 if x = (2k+1)π/2, k ∈ ℤ; 0 otherwise}</p><p><strong>Step 4:</strong> Check continuity and differentiability.</p><p>At x = (2k+1)π/2: lim(x→a) f(x) = 0 but f(a) = 1, so f is <strong>discontinuous</strong> at these points.</p><p>At all other points: f is continuous but has <strong>jump discontinuities</strong> nearby.</p><p>Since f is discontinuous at infinitely many points, it is <strong>not differentiable</strong> at those points.</p><p>∴ <strong>B:</strong> f is discontinuous at x = (2k+1)π/2, k ∈ ℤ</p><p>∴ <strong>D:</strong> f is not differentiable at x = (2k+1)π/2, k ∈ ℤ</p>
Correct Answer: BD