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?

Observe that when $x$ is very small near $0$, we have: 
If $x > 0 \Rightarrow 1+x < e^x < 1 + x + 2x^2 \to 1 < \dfrac{e^x-1}{x} < 1 + 2x$, and letting $x \to 0^{+}$, we have: $\displaystyle \lim_{x \to 0^{+}} \dfrac{e^x-1}{x} = 1 \tag{1}$,
and for $x < 0 \Rightarrow 1 > \dfrac{e^x-1}{x} > 1 + 2x$, and letting $x \to 0^{-}$, we have:
$\displaystyle \lim_{x\to 0^{-}} \dfrac{e^x-1}{x} = 1 \tag{2}$.
$(1),(2) \Rightarrow \displaystyle \lim_{x \to 0} \dfrac{e^x-1}{x} = 1$.
