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?
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