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?
 A: 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
$$
\lim_{x\to0}\frac{e^x-1}{x}=1\tag{6}
$$
A: 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.
A: $$e^x=1+x+1/2x^2+\cdots$$
$${e^x-1\over x}={x+x^2/2+\cdots\over x}=1+\frac12x+\cdots$$
$$\lim_{x\to0}{e^x-1\over x}=\lim_{x\to0}1+\frac12x+\cdots=1.$$
A: 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:
$\displaystyle \lim_{x\to 0^{-}} \dfrac{e^x-1}{x} = 1 \tag{2}$.
$(1),(2) \Rightarrow \displaystyle \lim_{x \to 0} \dfrac{e^x-1}{x} = 1$.
