How do you find the horizontal asymptote for (x^2+1)/(x^2-1)?
1 Answer
Jan 7, 2016
See explanation...
Explanation:
One way is to divide both numerator and denominator by
(x^2+1)/(x^2-1) = (1+1/x^2)/(1-1/x^2)
Then note that
So
(x^2+1)/(x^2-1) -> (1+0)/(1-0) = 1 asx->+-oo
So the horizontal asymptote is
Alternatively, separate out the "polynomial part"
(x^2+1)/(x^2-1) = (x^2-1+2)/(x^2-1) = 1 + 2/(x^2-1)
Then note that
So again we see that the horizontal asymptote is