How do you find the vertical, horizontal or slant asymptotes for #f(x)= (3e^(x))/(2-2e^(x))#?

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Answer 1 · 2016-04-16T09:30:54

Vertical asymptote is #x=0# and

two horizontal asymptotes #x=0# and #x=-3/2#

Vertical asymptote is given by putting denominator equal to zero.

Hence it is given by #2-2e^x=0# or #2e^x=2#

or #e^x=1# or #x=ln1=0#

For horizontal asymptote, let us find out #Lt_(x->oo)(3e^x)/(2-2e^x)#

Dividing numerator by #e^x# this is equal to

#Lt_(x->oo)3/(2e^-x-2)=-3/2# as #e^-oo=0#

Also #Lt_(x->-oo)(3e^x)/(2-2e^x)=0#

Hence we have two horizontal asymptotes #x=0# and #x=-3/2#

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