How to find the asymptotes of #f(x)= (3e^(x))/(2-2e^(x))#?

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Answer 1 · 2017-10-03T09:01:02

Horizontal asymptotes #y=0# and #y=1.5# and vertical asymptote #x=0#

One can find asymptotes of #f(x)# two ways

one , finding how #f(x)# behaves or tends as #x->oo# or #x->-oo#

and here as #f(x)=(3e^x)/(2-2e^x)=3/(2e^(-x)-2)#

as #x->-oo#, #e^x->0# and #f(x)->0/2=0#.

Similarly as #x->oo#, #e^(-x)->0# and #f(x)->3/2=1.5#

This gives us horizontal asymptotes as #y=0# and #y=1.5#

two as #f(x)=oo#, #2-2e^x=0# or #e^x=1# i.e. #x=0#

Hence we have a vertical asymptote #x=0#

graph{(3e^x)/(2-2e^x) [-10, 10, -5, 5]}