# How do you find all the asymptotes for function f(x)= (3e^(x))/(2-2e^(x))?

Jul 15, 2015

Vertical asymptote: $x = 0$
Horizontal asymptotes:
$y = 0$
$y = - \frac{3}{2}$

#### Explanation:

You start by checking which values of $x$ make your denominator equal to zero (you do not want this!).
To avoid zero in the denominator $x$ must be different from zero or:
$x \ne 0$ this means that the vertical line of equation $x = 0$ will be a "forbidden zone", i.e., a vertical asymptote.

To see if we have horizontal asymptotes we check the behaviour of your function for $x$ very large or, using the idea of Limit :

1] $x \to + \infty$
${\lim}_{x \to + \infty} \frac{3 {e}^{x}}{2 - 2 {e}^{x}} = {\lim}_{x \to + \infty} \frac{3 {e}^{x}}{{e}^{x} \left(\frac{2}{e} ^ x - 2\right)} =$
$= {\lim}_{x \to + \infty} \frac{3 \cancel{{e}^{x}}}{\cancel{{e}^{x}} \left(\frac{2}{e} ^ x - 2\right)} =$ and when $x \to + \infty$:
$= {\lim}_{x \to + \infty} \frac{3}{- 2} = - \frac{3}{2}$ (where I used the fact that $\frac{1}{e} ^ \infty = \frac{1}{\infty}$)
So, the horizontal asymptote will be the horizontal line of equation: $y = - \frac{3}{2}$.

2] $x \to - \infty$
${\lim}_{x \to - \infty} \frac{3 {e}^{x}}{2 - 2 {e}^{x}} = {\lim}_{x \to - \infty} \frac{3 {e}^{x}}{{e}^{x} \left(\frac{2}{e} ^ x - 2\right)} =$
$= {\lim}_{x \to - \infty} \frac{3 \cancel{{e}^{x}}}{\cancel{{e}^{x}} \left(\frac{2}{e} ^ x - 2\right)} =$ and when $x \to - \infty$:
$= {\lim}_{x \to - \infty} \frac{3}{\frac{2}{e} ^ x - 2} = 0$ (where I used the fact that $\frac{1}{e} ^ - \infty = {e}^{\infty}$).
So, the horizontal asymptote will be the horizontal line of equation: $y = 0$.

Graphically:
graph{(3e^x)/(2-2e^x) [-11.25, 11.25, -5.625, 5.625]}