Question

$\lim_{x \to 0} \frac{e^{ax} - e^{bx}}{x}=$

• A. a-b
• B. a+b
• C. ab²
• D. a³-b³

Answer 1

Answer:

A. a-b

$\lim_{x \to 0} \frac{e^{ax} - e^{bx}}{x}=$

Applying L Hospitals rule

$=\lim_{x \to 0} \frac{(e^{ax} \times a) - (e^{bx} \times b)}{1}=$

$=ae^0 - be^0$

$=a-b$