How do you find the horizontal asymptote for $f (x) = \frac{3 e^{x}}{2 - 2 e^{x}}$ $f(x)= (3e^(x))/(2-2e^(x))$ ?

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Answer 1 · 2016-01-11T13:33:52

$y = 0$ $y=0$
$y = - \frac{3}{2}$ $y=-3/2$

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]}

How do you find the horizontal asymptote for f(x)= (3e^(x))/(2-2e^(x))f(x)=3ex2−2exf(x)= (3e^(x))/(2-2e^(x))?