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

Jan 11, 2016

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

#### Explanation:

To find any horizontal asymptotes we have to evaluate:

${\lim}_{x \rightarrow \pm \infty} f \left(x\right)$

if the limits are finite then they are horizontal asymptotes

${\lim}_{x \rightarrow - \infty} f \left(x\right) = {\lim}_{x \rightarrow - \infty} \frac{3}{2} \left({e}^{x} / \left(1 - {e}^{x}\right)\right) =$

$\frac{3}{2} {\lim}_{x \rightarrow - \infty} \left({e}^{x} / \left(1 - {e}^{x}\right)\right) \approx \frac{3}{2} {\lim}_{x \rightarrow - \infty} {e}^{x} / 1 = \frac{3}{2} \cdot 0 = 0$

$\therefore y = 0$ is a horizontal asymptote for $x \rightarrow - \infty$

${\lim}_{x \rightarrow + \infty} f \left(x\right) = {\lim}_{x \rightarrow + \infty} \frac{3}{2} \left({e}^{x} / \left(1 - {e}^{x}\right)\right) =$

$\frac{3}{2} {\lim}_{x \rightarrow + \infty} \left({e}^{x} / \left(1 - {e}^{x}\right)\right) \approx \frac{3}{2} {\lim}_{x \rightarrow + \infty} \left({e}^{x} / - {e}^{x}\right) =$

$= \frac{3}{2} \cdot \left(- 1\right) = - \frac{3}{2}$

$\therefore y = - \frac{3}{2}$ is a horizontal asymptote for $x \rightarrow + \infty$

graph{(3/2)*((e^x)/(1-e^x)) [-10, 10, -5, 5]}