How do you find the asymptote(s) or holes of  f(x) = (x-1) / (x^2-1)?

May 18, 2016

We have a vertical asymptote at $x + 1 = 0$ and no horizontal or slanting asymptote. We have a hole at $x = 1$.

Explanation:

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

Although $\frac{1}{x + 1}$ may be defined for $x = 1$, as $x - 1$ has cancelled out, the fact is for $x = 1$, we do not have $\frac{x - 1}{{x}^{2} - 1}$. Hence, we have a hole a $x = 1$.

As we have $\left(x + 1\right)$ in denominator, we have a vertical asymptote $x + 1 = 0$.

As the degree of numerator is less than that of denominator, we do not have any horizontal or slanting asymptote.

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