<p>If \(f(x) = \lim_{n \to \infty} \frac{x^{2n}-1}{x^{2n}+1}\) then \(f(x)\) is discontinuous at</p>
<p>(a) \(x = 1\) only</p>
<p>(b) \(x = -1\) only</p>
<p>(c) \(x = 1\) and \(x = -1\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: The limit of x^(2n) depends critically on whether |x| < 1, |x| = 1, or |x| > 1. As n→∞, x^(2n) behaves differently in each region, creating a piecewise function with potential discontinuities at boundary points.
<p><strong>Step 1:</strong> Analyze the limit by cases based on the value of |x|.</p><p><strong>Case 1: |x| < 1</strong><br/>As n→∞, x^(2n)→0, so f(x) = (0-1)/(0+1) = -1</p><p><strong>Case 2: |x| = 1</strong><br/>If x = 1: f(1) = (1-1)/(1+1) = 0<br/>If x = -1: f(-1) = (1-1)/(1+1) = 0</p><p><strong>Case 3: |x| > 1</strong><br/>Divide numerator and denominator by x^(2n):<br/>f(x) = lim(n→∞) (1 - x^(-2n))/(1 + x^(-2n)) = (1-0)/(1+0) = 1</p><p><strong>Step 2:</strong> Construct the piecewise function:<br/>f(x) = { -1 if |x| < 1; 0 if |x| = 1; 1 if |x| > 1 }</p><p><strong>Step 3:</strong> Check continuity at x = 1 and x = -1.<br/>At x = 1: lim(x→1⁻) f(x) = -1, f(1) = 0, lim(x→1⁺) f(x) = 1 → discontinuous<br/>At x = -1: lim(x→-1⁻) f(x) = 1, f(-1) = 0, lim(x→-1⁺) f(x) = -1 → discontinuous</p><p>∴ Answer: f(x) is discontinuous at x = ±1 (or x = 1 and x = -1)</p>
Correct Answer: C