# What are the asymptote(s) and hole(s), if any, of  f(x) =(1-x)^2/(x^2-1)?

##### 1 Answer
Nov 21, 2015

$f \left(x\right)$ has a horizontal asymptote $y = 1$, a vertical asymptote $x = - 1$ and a hole at $x = 1$.

#### Explanation:

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

$= 1 - \frac{2}{x + 1}$

with exclusion $x \ne 1$

As $x \to \pm \infty$ the term $\frac{2}{x + 1} \to 0$, so $f \left(x\right)$ has a horizontal asymptote $y = 1$.

When $x = - 1$ the denominator of $f \left(x\right)$ is zero, but the numerator is non-zero. So $f \left(x\right)$ has a vertical asymptote $x = - 1$.

When $x = 1$ both the numerator and denominator of $f \left(x\right)$ are zero, so $f \left(x\right)$ is undefined and has a hole at $x = 1$. Note that ${\lim}_{x \to 1} f \left(x\right) = 0$ is defined. So this is a removable singularity.