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

Jun 27, 2016

right sided asymptote is $y = - \frac{3}{2}$

left sided asymptote is $y = 0$

#### Explanation:

for horizontal asymptotes we should look at

${\lim}_{x \setminus \to \pm \infty} \frac{3 {e}^{x}}{2 - 2 {e}^{x}}$

$= {\lim}_{x \setminus \to \pm \infty} \frac{3}{2 {e}^{- x} - 2}$

$= {\lim}_{x \setminus \to \pm \infty} \frac{\frac{3}{2}}{{e}^{- x} - 1}$

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

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

FOR $x \setminus \to + \infty$

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

so ${\lim}_{x \setminus \to + \infty} \frac{\frac{3}{2}}{{e}^{- x} - 1} = - \frac{3}{2}$

but FOR $x \setminus \to - \infty$

${e}^{- x} {|}_{x \setminus \to - \infty}$

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

$= \infty$

so ${\lim}_{x \setminus \to - \infty} \frac{\frac{3}{2}}{{e}^{- x} - 1} = 0$