Question

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

Answer 1

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 `(x^2-1)/(x^2-1)=x^2/x^2=1`
`x->oo`
graph{(y-(x^2+1)/(x^2-1))(y-1)=0 [-7.9, 7.9, -3.95, 3.95]}