Question:

Evaluate: \[ \lim_{x \to 0} \left[ \frac{1}{x} - \frac{1}{e^x - 1} \right] \]

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Use Taylor expansion to approximate functions near \( x = 0 \) to evaluate limits.
Updated On: Mar 24, 2025
  • \( 0 \)
  • \( 1 \)
  • \( 2 \)
  • \( \frac{1}{2} \)
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The Correct Option is D

Solution and Explanation

Step 1: Using Taylor series expansion Using \( e^x \approx 1 + x + \frac{x^2}{2} \) for small \( x \), we approximate: \[ e^x - 1 \approx x + \frac{x^2}{2} \]
Step 2: Simplifying the expression \[ \frac{1}{x} - \frac{1}{e^x - 1} = \frac{e^x - 1 - x}{x(e^x - 1)} \] Substituting the approximation, \[ \frac{x + \frac{x^2}{2} - x}{x(x + \frac{x^2}{2})} = \frac{\frac{x^2}{2}}{x^2 + \frac{x^3}{2}} \]
Step 3: Evaluating the limit Taking \( x \to 0 \), we get: \[ \frac{1}{2} \]
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