# What are the asymptote(s) and hole(s), if any, of  f(x) = x/(x-1)-(x-1)/x?

##### 1 Answer
Jun 12, 2017

$x = 0$ is an asymptote.

$x = 1$ is an asymptote.

#### Explanation:

First, let's simplify this so that we have a single fraction that we can take the limit of.

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

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

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

Now, we need to check for discontinuities. This is just anything that will make the denominator of this fraction $0$. In this case, to make the denominator $0$, $x$ could be $0$ or $1$. So let's take the limit of $f \left(x\right)$ at those two values.

${\lim}_{x \to 0} \frac{2 x - 1}{x \left(x - 1\right)} = \frac{- 1}{- 1 \cdot 0} = \pm \infty$

${\lim}_{x \to 1} \frac{2 x - 1}{x \left(x - 1\right)} = \frac{3}{1 \cdot 0} = \pm \infty$

Since both of these limits tend towards infinity, both $x = 0$ and $x = 1$ are asymptotes of the function. There are therefore no holes in the function.