I am finding (and computing the limits) of all vertical and horizontal asymptotes of ${5(e^x) \over (e^x)-2}$. I have found the vertical asymptote at $x=ln2$. I have also found the horizontal asymptote at y=5 by calculating the limit of the function at positive infinity. I know that to find the other horizontal asymptote (which is at $y=0$, based on the graph), I need to calculate the limit at negative infinity, but I cannot get it to equal zero.
my calculations thus far: lim as x approaches negative infinity of the function is equal to the limit as x approaches infinity of ${5 \over (e^x)-2}$, which I found by dividing both numerator and denominator by $e^x$.
but I just keep getting that the limit at negative infinity is the same as the limit at positive infinity - ${5\over(1-0)}$ or 5 - which I know is not correct.
what am i missing?