how in the world do I find the asymptope of (e^x)/(1+e^x) i have no idea other than lim x->oo will come out with the verticle and lim x->o+ will comeout with the horizontal. ultimately i got y=1 and x=0 but is that correct?
2006-07-24 15:15:35 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Well, the limit as x->0+ is 0.5. The limit as x->inf is 1. I don't know those are the asymptotes, but I know they are the limits you seek.
2006-07-24 15:21:12 · answer #1 · answered by anonymous 7 · 0⤊ 0⤋
The asymptotes are y=0 and y=1.
You first consider the limit of your function as x approaches positive infinity (it's 1) and then as x approaches negative infinity (it's 0). These tell you what your asymptotes will be.
2006-07-24 22:26:13 · answer #2 · answered by Anonymous · 0⤊ 0⤋
when denominator = 0, u find an asymptote.
1+e^x = 0 ==> e^x = -1 brcause e^x can not be equal to zero, we canot find an asymptote.
as x==> + infinite, f(x) = (e^x)/(1+e^x) divide by e^x, f(x) = 1/(1/e^x + 1)= 1/(0 + 1) = 1. so as x==> + infinite, y==>1 so the asymtote is y=1
as x==> - infinite, f(x) = (e^x)/(1+e^x) divide by e^x, f(x) = 1/(1/e^x + 1)= 1/(infinite + 1) = 0. so as x==> - infinite, y==>0 so the asymtote is y=0
totally there is two asymptotes
2006-07-24 22:35:35 · answer #3 · answered by ___ 4 · 0⤊ 0⤋
I would think y=1 and y=0 are the asymptotes
refer figure on this page
http://qwpey.com/mathindex/gl/tra.php
2006-07-24 22:30:52 · answer #4 · answered by qwert 5 · 0⤊ 0⤋