Question

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

• A. 1
• B. e
• C. -1
• D.

Answer 1

The correct option is A

1

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

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

We know, expression of $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+........$

$\Rightarrow \underset{x\to 0}{lim}\frac{(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+......)-1}{x}$

$=\underset{x\to 0}{lim}\frac{(x+\frac{x^2}{2!}+\frac{x^3}{3!}+......)}{x}$

$=\underset{x\to 0}{lim}(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+......)$

$=1+0$

$=1$

Option A is correct