Question

The value of $ \lim_{x \to 0} \frac{e^{ax} - e^{bx}}{2x} \text{ is equal to} $

• A. \( \frac{a + b}{2} \)
• B. \( \frac{a - b}{2} \)
• C. \( \frac{e^{ab}}{2} \)
• D. \( 0 \)

Hint:

For limits involving small values of \( x \), use the first-order approximation \( e^x \approx 1 + x \) when \( x \to 0 \).

Answer 1

The Correct Option is B

We are given the limit expression: \[ \lim_{x \to 0} \frac{e^{ax} - e^{bx}}{2x} \] Using the first-order approximation of the exponential function for small values of \( x \), we have: \[ e^{ax} \approx 1 + ax \quad \text{and} \quad e^{bx} \approx 1 + bx \quad \text{as} \quad x \to 0. \] Substitute these approximations into the limit expression: \[ \frac{e^{ax} - e^{bx}}{2x} \approx \frac{(1 + ax) - (1 + bx)}{2x} = \frac{ax - bx}{2x} = \frac{(a - b)x}{2x}. \] Simplifying: \[ \frac{a - b}{2}. \] Thus, the value of the given limit is \( \frac{a - b}{2} \).