Prove that $\lim_\limits{x\to 0}{\frac{e^x-1}{x}}=1$ without derivatives 
Prove that $\lim_\limits{x\to 0}{\frac{e^x-1}{x}}=1$.

I currently know only one approach (using L'Hopital 's Rule and derivatives) as follows:
$$\lim_\limits{x\to 0}{\frac{e^x-1}{x}}=\lim_\limits{x\to 0}{\frac{\left(e^x-1\right)'}{x'}}=\lim_\limits{x\to 0}{\left(e^x\right)}=e^0=1$$
Here I ask for other proofs than those, preferably neither using derivatives in any way nor using Taylor, etc.
For the purposes of this post, I define the exponential by any of the following limits:
$$e^x
=\lim_\limits{n\to +\infty}{\left( 1+\frac{x}{n}\right)^n}=\lim_\limits{n\to +\infty}{\left[ \left( 1+\frac{1}{n}\right)^{n\cdot x}\right] }=\left[ \lim_\limits{n\to +\infty}{\left( 1+\frac{1}{n}\right)^n}\right] ^x.$$
Note: An approach for $\lim_\limits{x\to 0^+}{(x\ln x)}$ without using derivatives can be found here.
 A: We can assume the "Important limit" $\lim_{x\rightarrow 0}(1+x)^{1/x}=e$. Then
\begin{align*}
\lim_{x\rightarrow 0} \frac{e^x-1}{x} &= \lim_{y\rightarrow 0}\frac{y}{\ln (1+y)}\\
&= \lim_{y\rightarrow 0}\frac{1}{\ln \left((1+y)^{1/y}\right)}\\
&= 1.
\end{align*}
A: Using 
$$e^x=\lim_{n\to\infty}(1+\frac{x}{n})^n=1+x+O(x^2)$$ 
Which is not directly using the Taylor Series, just the binomial expansion.
You can procede like in the proof of using the Taylor series.
A: Since $1 + x \le e^x$ for all $x$ and we have $e^x \le 1/(1 - x)$ for $x < 1$
so 
$$1 \le (e^x - 1)/x \le 1/(1 - x) \to 1$$ as $x\to 0$.
A: Using that
$$\exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$$
and that the series
$$\frac{e^x-1}{x}=\sum_{n\ge0}\frac{x^n}{(n+1)!}$$
is uniformly convergent on every interval containing $0$ so interchanging limit and $\sum$ is allowed and the result follows.
A: If you are allowed to use the fact that $e^z$ is analytic and hence you can take $z=iy\rightarrow 0$ and obtain the same limit (assuming it exists) the  you get $\frac{\sin y}{y}$ whose limit can be obtained from geometrically.
A: Using the definition of e, the limit becomes the same as:
$$\lim_{x\to 0}\lim_{n\to \infty} \frac{
\left(1+\frac{1}{n}\right)^{nx}-1}{x}$$
We can evaluate this along the path $n=\frac{1}{x}$. So, the limit becomes:
$$\lim_{x\to 0} \frac{
\left(1+x\right)^{x/x}-1}{x}$$
This gives us:
$$ \lim_{x\to 0} \frac{
\left(1+x\right)-1}{x} = 1$$
A: maybe this satisfies. 
$$
\frac{e^x-1}{x} = e^{x/2}\frac{e^{x/2}-e^{-x/2}}{x} \to  e^{it}\frac{e^{it}-e^{-it}}{i2t} = \frac{\sin t}{t}e^{it}
$$
Then taking limits to zero. 
