Hello,

The function is

$\displaystyle y = \frac{e^x + e^{-x}}{e^x - e^{-x}}$

I managed to get to the first horizontal asymptote by using lhospital's rule after I got to this:

$\displaystyle y = \frac{e^{2x} + 1}{e^{2x} - 1} = \frac{(e^{2x} + 1)'}{(e^{2x} - 1)'} = \frac{e^{2x} * 2}{e^{2x} * 2} = \frac{1}{1} = 1$

But then the sign of infinity which I insert doesn't matter because y will always be 1, and I know that there's another asymptote in y=-1.

How do I get to the other asymptote? Have I done anything wrong?