Question

The value of $\lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)$ is

(a) $\frac{1}{2}$

(b) $\infty$

(c) 1

(d) 0

Answer 1

$\lim _{x \rightarrow 0} \frac{e^x-1}{x}=\lim _{x \rightarrow 0}\left(\frac{1+\frac{x}{1 !}+\frac{x^2}{2 !}+\ldots-1}{x}\right)$

$=\lim _{x \rightarrow 0}\left(1+\frac{x}{2 !}+\frac{x^2}{3 !}+\ldots \infty\right)=1$

So, The correct option of this question will be (C).