How do you find the horizontal asymptote for #f(x)= (3e^(x))/(2-2e^(x))#?

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Answer 1 · 2016-01-11T13:33:52

#y=0#
#y=-3/2#

To find any horizontal asymptotes we have to evaluate:

#lim_(x rarr+-oo)f(x)#

if the limits are finite then they are horizontal asymptotes

#lim_(x rarr-oo)f(x)=lim_(x rarr-oo)3/2(e^x/(1-e^x))=#

#3/2lim_(x rarr-oo)(e^x/(1-e^x))~~3/2lim_(x rarr-oo)e^x/1=3/2*0=0#

#:. y=0# is a horizontal asymptote for #x rarr-oo#

#lim_(x rarr+oo)f(x)=lim_(x rarr+oo)3/2(e^x/(1-e^x))=#

#3/2lim_(x rarr+oo)(e^x/(1-e^x))~~3/2lim_(x rarr+oo)(e^x/-e^x)=#

#=3/2*(-1)=-3/2#

#:. y=-3/2# is a horizontal asymptote for #x rarr+oo#

graph{(3/2)*((e^x)/(1-e^x)) [-10, 10, -5, 5]}