<p>Evaluate: <span>\(\lim_{x \to 0} \dfrac{1}{x} - \dfrac{2}{e^{2x} - 1}\)</span></p>
Step-by-Step Solution
Key Concept: Rewrite the expression with a common denominator and use Taylor expansion of e^(2x) around x=0 to resolve the indeterminate form. The key is recognizing that both terms have competing singularities that nearly cancel.
<p><strong>Step 1:</strong> Combine the fractions with common denominator x(e^(2x) - 1):</p><p>$$\lim_{x \to 0} \frac{1}{x} - \frac{2}{e^{2x} - 1} = \lim_{x \to 0} \frac{e^{2x} - 1 - 2x}{x(e^{2x} - 1)}$$</p><p><strong>Step 2:</strong> Expand e^(2x) using Taylor series: e^(2x) = 1 + 2x + 2x² + \frac{4x³}{3} + ...</p><p>$$e^{2x} - 1 = 2x + 2x² + \frac{4x³}{3} + ...$$</p><p><strong>Step 3:</strong> Numerator: e^(2x) - 1 - 2x = 2x² + \frac{4x³}{3} + ...</p><p><strong>Step 4:</strong> Denominator: x(e^(2x) - 1) = x(2x + 2x² + \frac{4x³}{3} + ...) = 2x² + 2x³ + ...</p><p><strong>Step 5:</strong> Evaluate the limit:</p><p>$$\lim_{x \to 0} \frac{2x² + \frac{4x³}{3} + ...}{2x² + 2x³ + ...} = \lim_{x \to 0} \frac{2x²(1 + \frac{2x}{3} + ...)}{2x²(1 + x + ...)} = \frac{1}{1} = 1$$</p><p>∴ Answer: <strong>D (1)</strong></p>
Correct Answer: D