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?

I don't know if this is really "without" Hôpitals rule for you but if you are allowed to use
$$
\exp(x)=\sum_{k=0}^\infty \frac{x^k}{k!}
$$
the limit is straightforward since this sum converges locally uniformly.
