How do you find all the asymptotes for function #f(x)= (3e^(x))/(2-2e^(x))#?

1 Answer
Jul 15, 2015

Answer:

Vertical asymptote: #x=0#
Horizontal asymptotes:
#y=0#
#y=-3/2#

Explanation:

You start by checking which values of #x# make your denominator equal to zero (you do not want this!).
To avoid zero in the denominator #x# must be different from zero or:
#x!=0# this means that the vertical line of equation #x=0# will be a "forbidden zone", i.e., a vertical asymptote.

To see if we have horizontal asymptotes we check the behaviour of your function for #x# very large or, using the idea of Limit :

1] #x->+oo#
#lim_(x->+oo) (3e^x)/(2-2e^x)=lim_(x->+oo) (3e^x)/(e^x(2/e^x-2))=#
#=lim_(x->+oo) (3cancel(e^x))/(cancel(e^x)(2/e^x-2))=# and when #x->+oo#:
#=lim_(x->+oo) 3/(-2)=-3/2# (where I used the fact that #1/e^oo=1/oo#)
So, the horizontal asymptote will be the horizontal line of equation: #y=-3/2#.

2] #x->-oo#
#lim_(x->-oo) (3e^x)/(2-2e^x)=lim_(x->-oo) (3e^x)/(e^x(2/e^x-2))=#
#=lim_(x->-oo) (3cancel(e^x))/(cancel(e^x)(2/e^x-2))=# and when #x->-oo#:
#=lim_(x->-oo) 3/(2/e^x-2)=0# (where I used the fact that #1/e^-oo=e^oo#).
So, the horizontal asymptote will be the horizontal line of equation: #y=0#.

Graphically:
graph{(3e^x)/(2-2e^x) [-11.25, 11.25, -5.625, 5.625]}