# Thread: Horizontal asymptotes / infinite limits and natural log

1. ## Horizontal asymptotes / infinite limits and natural log

Hi. This problem simply involves finding the horizontal asymptotes of the following function. While the calculator shows two (y = 0 when x approaches negative infinity and y = 2 for positive infinity), I can only come to the conclusion that y = 2, regardless of whether X approaches positive or negative infinity.

Function:
Y = (2e^x) / ((e^x)-5)

What needs to be done in order to get the result that Y approaches 0 as X approaches negative infinity? Much thanks in advance.

2. Originally Posted by kylera
Hi. This problem simply involves finding the horizontal asymptotes of the following function. While the calculator shows two (y = 0 when x approaches negative infinity and y = 2 for positive infinity), I can only come to the conclusion that y = 2, regardless of whether X approaches positive or negative infinity.

Function:
Y = (2e^x) / ((e^x)-5)

What needs to be done in order to get the result that Y approaches 0 as X approaches negative infinity? Much thanks in advance.
$y = \frac{2 e^x}{e^x - 5} = \frac{2}{1 - 5 e^{-x}}$.

$\lim_{x \rightarrow + \infty} \frac{2}{1 - 5 e^{-x}} = \frac{2}{1 - 5(0)} = 2$.

$\lim_{x \rightarrow - \infty} \frac{2 e^x}{e^x - 5} = \frac{0}{0 - 5} = 0$.

The two horiziontal asymptotes are therefore y = 2 and y = 0.

3. Stupid oversight on my part. Much thanks again.