Question

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

• A. $\frac{1}{2}$
• B. $\infty$
• C. $1$
• D. $0$

Answer 1

The correct option is C $1$

${lim}_{x\to 0}\left(\frac{e^x-1}{x}\right)$
$={lim}_{x\to 0}\frac{(1+\frac{x}{1!}+\frac{x^2}{2!}+.......-1)}{x}$
${lim}_{x\to 0}(1+\frac{x}{2!}+\frac{x^2}{3!}+.......+\infty )=1$