Limits, Continuity & Differentiability
Continuity
Grade None
Question:
<p>Given \(f(x) = \dfrac{1}{x} - \dfrac{k-1}{e^{2x}-1},\; x \neq 0\). If \(f(x)\) is continuous at \(x = 0\) and \(f(0) = 1\), find the value of \(k\).</p>
<p>\(1\)</p>
<p>\(3\)</p>
<p>\(2\)</p>
<p>\(4\)</p>
Step-by-Step Solution
Key Concept: For continuity at x=0, we need lim(x→0) f(x) = f(0) = 1. Both fractions have indeterminate forms as x→0, so we must use Taylor series expansions: 1/x has a pole, but the second term must cancel it while the limit equals 1.
<p><strong>Step 1:</strong> For continuity at x=0, we need lim(x→0) f(x) = f(0) = 1.</p><p><strong>Step 2:</strong> Expand using Taylor series near x=0:</p><p>• e^(2x) = 1 + 2x + 2x² + (4x³)/3 + ...</p><p>• e^(2x) - 1 = 2x + 2x² + (4x³)/3 + ...</p><p>• 1/(e^(2x)-1) = 1/(2x(1 + x + (2x²)/3 + ...)) = (1/2x) · 1/(1 + x + (2x²)/3 + ...)</p><p>• 1/(e^(2x)-1) = (1/2x)(1 - x + (x²)/3 + ...) = 1/(2x) - 1/2 + x/6 + ...</p><p><strong>Step 3:</strong> Substitute into f(x):</p><p>f(x) = 1/x - (k-1)[1/(2x) - 1/2 + x/6 + ...]</p><p>f(x) = 1/x - (k-1)/(2x) + (k-1)/2 - (k-1)x/6 + ...</p><p><strong>Step 4:</strong> For the 1/x terms to cancel: 1 - (k-1)/2 = 0</p><p>→ (k-1)/2 = 1 → k-1 = 2 → k = 3</p><p><strong>Step 5:</strong> Verify the limit with k=3:</p><p>f(x) = (3-1)/2 - (3-1)x/6 + ... = 1 - x/3 + ...</p><p>lim(x→0) f(x) = 1 = f(0) ✓</p><p>∴ Answer: <strong>B (k = 3)</strong></p>
Correct Answer: B