Question

Prove that : $\underset{x\to 0}{lim}\left(\frac{e^x-1}{x}\right)=1$

Answer 1

$L.H.S.=\underset{x\to 0}{lim}\left(\frac{e^x-1}{x}\right)$
$=\underset{x\to 0}{lim}\left[\frac{1}{x}\{(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...)-1\}\right]$
$=\underset{x\to 0}{lim}\left[\frac{1}{x}(x+\frac{x^2}{2!}+\frac{x^3}{3!}+....)\right]$
$=\underset{x\to 0}{lim}[1+\frac{x}{2!}+\frac{x^2}{3!}+....]$
$=1+0+0+......$
$=1=R.H.S.$
$L.H.S.=R.H.S.$