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

To solve the problem, we need to evaluate the limit:

\[ \lim_{x \to 0} \frac{e^{ax} - e^{bx}}{2x} \]

1. Expand the Exponential Functions:
Using the Taylor series expansion for exponential functions:

\[ e^{ax} = 1 + ax + \frac{(ax)^2}{2!} + \frac{(ax)^3}{3!} + \cdots \]
\[ e^{bx} = 1 + bx + \frac{(bx)^2}{2!} + \frac{(bx)^3}{3!} + \cdots \]

2. Substitute the Expansions:
Substitute these expansions into the original limit expression:

\[ \lim_{x \to 0} \frac{\left( 1 + ax + \frac{(ax)^2}{2!} + \cdots \right) - \left( 1 + bx + \frac{(bx)^2}{2!} + \cdots \right)}{2x} \]

3. Simplify the Numerator:
Combine like terms in the numerator:

\[ \lim_{x \to 0} \frac{x \left[ (a - b) + x\left( \frac{a^2}{2!} - \frac{b^2}{2!} \right) + x^2\left( \frac{a^3}{3!} - \frac{b^3}{3!} \right) + \cdots \right]}{2x} \]

4. Cancel the Common Factor:
Cancel \( x \) from numerator and denominator:

\[ \lim_{x \to 0} \frac{1}{2} \left[ (a - b) + x\left( \frac{a^2 - b^2}{2} \right) + x^2\left( \frac{a^3 - b^3}{6} \right) + \cdots \right] \]

5. Evaluate the Limit:
As \( x \to 0 \), all terms containing \( x \) vanish, leaving:

\[ \frac{a - b}{2} \]

Final Answer:
The value of the limit is \( \boxed{\dfrac{a - b}{2}} \).

Answer 2

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 \( \frac{a - b}{2} \).