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

1 Answer
Nov 5, 2016

Answer:

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]}