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?
Best Answer
Let $y=-x$; then you can more easily see that
$$\lim_{x\to-\infty}e^x=\lim_{y\to+\infty}e^{-y}=\lim_{y\to+\infty}\frac1{e^y}=0\;,$$
since the denominator $e^y$ is blowing up rather explosively as $y\to\infty$.
Going back to the original problem,
$$\lim_{x\to-\infty}\frac{5e^x}{e^x-2}=\frac{\lim\limits_{x\to-\infty}5e^x}{\lim\limits_{x\to-\infty}e^x-2}=\frac0{0-2}=0\;.$$