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

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Answer 1 · 2016-06-27T21:57:57

right sided asymptote is $y = - 3/2$

left sided asymptote is $y = 0$

for horizontal asymptotes we should look at

$lim_{x \to pm oo} (3e^x)/(2-2e^x)$

$= lim_{x \to pm oo} (3)/(2e^{-x}-2)$

$= lim_{x \to pm oo} (3/2)/(e^{-x}-1)$

which we get by having divided top and bottom by $e^x$

So we focus on what happens to $e^{-x}$ as $x \to pm oo$

FOR $x \to + oo$

well, $e^{-x} |_{x \to + oo} = 0$

so $lim_{x \to + oo} (3/2)/(e^{-x}-1) = - 3/2$

but FOR $x \to - oo$

$e^{-x} |_{x \to - oo}$

$= e^{x} |_{x \to + oo}$

$= oo$

so $lim_{x \to - oo} (3/2)/(e^{-x}-1) = 0$

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