Question

What are the equations of the horizontal asymptotes of y = (4e^x+7)/(e^x-1) in the xy plane?

Answer 1

\[y = \frac{ 4e^x + 7}{ e^x - 1 }\]

Answer 2

I know one horizontal asymptote is y = 4. But there are two...
The other one is y = -7, but how do you get that one?
I know you can break up the numerator so that
\[y = \frac{ 4e^x }{ e^x - 1 } + \frac{ 7 }{ e^x - 1 }\]
but how would you get the -7?

Answer 3

as \(x\to -\infty\) you have \(e^x\to 0\)

Answer 4

ohhhhh i see it now.
-_- wow. okay, thanks.

Answer 5

so
\[\lim_{x\to -\infty}\frac{ 4e^x + 7}{ e^x - 1 }=\frac{7}{-1}\]

Answer 6

yw