There is a very simple limit that I can prove using L'Hôpital's rule and other techniques, but I can't really do it using the definition of $\epsilon-\delta$.
$$ \lim_{x\rightarrow 0}\frac{e^{x} -1}{x} =1 $$
Please help! I've been stuck with this for a few weeks. I've even asked some old teachers of mine, and nobody was able to solve it.
Thanks in advance!
EDIT
As asked, I'm going to show you how far I got. I'm sure I made mistakes, but like I said, I'm really at a loss with this limit. Thanks.
$$ \lim _{x\rightarrow 0}\frac{e^{x} -1}{x} =1\ \ \Longleftrightarrow \ \ \forall \epsilon >0\ \ \exists \delta >0\ \ ||\ \ 0< |x|< \delta \ \ \Rightarrow \ \ \left| \frac{e^{x} -1}{x} -1\right| < \epsilon $$
$$ \left| \frac{e^{x} -1}{x}\right| -| 1| \leqslant \left| \frac{e^{x} -1}{x} -1\right| < \epsilon \ \ \Rightarrow \ \ \left| \frac{e^{x} -1}{x}\right| < \epsilon +1 $$
$$ \frac{\left| e^{x} -1\right|}{| x| } = \left| \frac{e^{x} -1}{x}\right| < \epsilon +1 $$
If $\delta \leqslant 1$ then:
$$ -1< x< 1 $$ $$ e^{-1} < e^{x} < e $$ $$ e^{-1} -1< e^{x} -1< e-1 $$ $$ -( e-1) < e^{-1} -1< e^{x} -1< e-1 $$ $$ \left| e^{x} -1\right| < e-1 $$
So now:
$$ \frac{\left| e^{x} -1\right| }{| x| } < \frac{e-1}{| x| } < \epsilon +1\ \ \Longrightarrow \ \frac{e-1}{\epsilon +1} < | x| < \delta < 1 $$
As you can see, the last part is a contradiction, so clearly there's something I'm doing wrong here.
Thanks.