# How do you find the asymptotes for (x^2+1)/(x^2-1)?

##### 1 Answer
Nov 5, 2016

The vertical asymptotes are $x = 1$ and $x = - 1$
The horizontal asymptote is $y = 1$

#### Explanation:

The denominator cannot be divided by $0$.
So the vertical asymptotes are $x = 1$ and $x = - 1$
As the degree of the numerator and denominator are the same, we would not expect an oblique asymptote.
Limit $\frac{{x}^{2} - 1}{{x}^{2} - 1} = {x}^{2} / {x}^{2} = 1$
$x \to \infty$
graph{(y-(x^2+1)/(x^2-1))(y-1)=0 [-7.9, 7.9, -3.95, 3.95]}