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QUESTION

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

y=0 y=-3/2

To find any horizontal 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]}