Question

$\underset{x\to 0}{lim}\frac{e^{\alpha x}-e^{\beta x}}{x}=$

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$\underset{x\to 0}{lim}\frac{e^{\alpha x}-e^{\beta x}}{x}$

At x=0, the value of the given function is $\frac{0}{0}$ form

$\underset{x\to 0}{lim}\frac{e^{\alpha x}-e^{\beta x}+1-1}{x}$

$\underset{x\to 0}{lim}\frac{(e^{\alpha x}-1)-(e^{\beta x}-1)}{x}$

$\underset{x\to 0}{lim}\frac{e^{\alpha x}-1}{x}\times \frac{\alpha }{\alpha }-\underset{x\to 0}{lim}\frac{e^{\beta x}-1}{x}\times \frac{\beta }{\alpha }$

$\alpha \underset{x\to 0}{lim}\frac{e^{\alpha x}-1}{\alpha x}-\beta \underset{x\to 0}{lim}\frac{e^{\beta x}-1}{\beta x}\{\underset{x\to 0}{lim}\frac{e^x-1}{x}=1\}$

$\alpha \times 1-\beta \times 1$

$\alpha -\beta$

Option D is correct