# For what values of x, if any, does f(x) = x/(e^x-e^(2x))  have vertical asymptotes?

Jan 9, 2017

The function $\frac{x}{{e}^{x} - {e}^{2 x}}$ does not have vertical asymptotes.

#### Explanation:

If we write the function as:

$f \left(x\right) = \frac{x}{{e}^{x} - {e}^{2 x}} = \frac{x}{{e}^{x} \left(1 - {e}^{x}\right)}$

we can easily see that $f \left(x\right)$ is continuous in all of $\mathbb{R}$ except for the point $x = 0$ where the denominator vanishes.

Analysing the limit:

${\lim}_{x \to 0} f \left(x\right) = {\lim}_{x \to 0} \frac{x}{{e}^{x} \left(1 - {e}^{x}\right)} = {\lim}_{x \to 0} \frac{1}{e} ^ x \cdot - \frac{1}{{\lim}_{x \to 0} \frac{{e}^{x} - 1}{x}} = 1 \cdot - \frac{1}{1} = - 1$

$\lim \frac{{e}^{x} - 1}{x} = 1$

So the function does not have vertical asymptotes.

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