Question

The value of $\underset{x\to 0}{\lim }\frac{\left[e^{ax}-e^{bx}\right]}{x}$ is

• A. $a+b$
• B. $a-b$
• C. $e^{ab}$
• D. $1$

Answer 1

The correct option is B

$a-b$

Explanation of the correct option.

Compute the required value.

Given : $\underset{x\to 0}{\lim }\frac{\left[e^{ax}-e^{bx}\right]}{x}$

$$\begin{gathered} =\frac{1-1}{0} \\ =\frac{0}{0} \end{gathered}$$

Since it is $\frac{0}{0}$ form, apply the L.Hospital rule.

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{a.e^{ax}-b.e^{bx}}{1} \\ =a-b \end{gathered}$$

Therefore the value of $\underset{x\to 0}{\lim }\frac{\left[e^{ax}-e^{bx}\right]}{x}$ is $a-b$.

Hence,option $\left(B\right)$ is the correct option.