Calculate $\lim_{x \to 0} (e^x-1)/x$ without using L'Hôpital's rule Any ideas on how to calculate the limit of $(e^x -1)/{x}$ as $x$ goes to zero without applying L'Hôpital's rule?

$$e^x=1+x+1/2x^2+\cdots$$
$${e^x-1\over x}={x+x^2/2+\cdots\over x}=1+\frac12x+\cdots$$
$$\lim_{x\to0}{e^x-1\over x}=\lim_{x\to0}1+\frac12x+\cdots=1.$$
