limits - $lim _{xto -infty }left(frac{left(e^x-1right)}{left(e^{2x}+1right)}right)$

How can I calculate the following limit?

$\lim _{x\to -\infty }\left(\frac{\left(e^x-1\right)}{\left(e^{2x}+1\right)}\right)$

If the limit is

$\lim _{x\to +\infty }\left(\frac{\left(e^x-1\right)}{\left(e^{2x}+1\right)}\right)$

then it is quiet easy, as I just need to make something like

$\lim _{x\to +\infty }\left(\frac{e^x\left(1-\frac{1}{e^x}\right)}{e^{2x}\left(1+\frac{1}{e^{2x}}\right)}\right)$

and it evident it is 0.

But with limit to negative infinity, I cannot do the same, as a I go back to an undetermined form, like

$\frac{0*\infty}{0*\infty}$

So don't know what I should do. Any suggestion?