TITLE: Calculate $\lim_{x \to 0} (e^x-1)/x$ without using L'Hôpital's rule
QUESTION [5 upvotes]: Any ideas on how to calculate the limit of $(e^x -1)/{x}$ as $x$ goes to zero without applying L'Hôpital's rule?

REPLY [11 votes]: Using Bernoulli's Inequality, for all $x$ so that $|x|\le n$,
$$
1+x\le\left(1+\frac xn\right)^n\tag{1}
$$
Therefore, letting $n\to\infty$, we get for all $x$,
$$
1+x\le e^x\tag{2}
$$
Furthermore, for $|x|\lt1$,
$$
1-x\le e^{-x}\implies\frac1{1-x}\ge e^x\tag{3}
$$
Thus, subtracting $1$ from $(2)$ and $(3)$ gives
$$
x\le e^x-1\le\frac x{1-x}\tag{4}
$$
Since we are looking for the limit as $x\to0$, assume that $|x|\lt1$. Whether $x$ is positive or negative, $(4)$ says that
$$
\frac{e^x-1}{x}\text{ is between $1$ and }\frac1{1-x}\tag{5}
$$
Therefore, by the Squeeze Theorem, we get
$$
\lim_{x\to0}\frac{e^x-1}{x}=1\tag{6}
$$