Question

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

• A. \( 0 \)
• B. \( 1 \)
• C. \( 2 \)
• D. \( \frac{1}{2} \)

Hint:

Use Taylor expansion to approximate functions near \( x = 0 \) to evaluate limits.

Answer 1

The Correct Option is D

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: \[ \frac{1}{2} \]

Answer 2

To evaluate the limit \(\lim_{x \to 0} \left[ \frac{1}{x} - \frac{1}{e^x - 1} \right]\), we begin by considering the behavior of each term as \(x\) approaches 0. We can use the first-order Taylor expansion for \(e^x\) around 0: \(e^x = 1 + x + \frac{x^2}{2} + \cdots\).
Thus, \(e^x - 1 \approx x\) when \(x\) is close to 0.
Now, recognize the expression: \(\frac{1}{x} - \frac{1}{e^x - 1} = \frac{1}{x} - \frac{x}{(e^x - 1)x} = \frac{(e^x - 1) - x}{x(e^x - 1)}\).
Simplifying the numerator, \((e^x - 1) - x \approx 1 + x + \frac{x^2}{2} - 1 - x = \frac{x^2}{2}\).
Therefore, the expression becomes:
\(\frac{\frac{x^2}{2}}{x \cdot x} = \frac{x}{2}\).
Hence, as \(x \to 0\), the expression approaches \(\frac{x}{2x} = \frac{1}{2}\).
Therefore, the limit is:

\(\lim_{x \to 0} \left[ \frac{1}{x} - \frac{1}{e^x - 1} \right] = \frac{1}{2}\).