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 to find:

(x^2+1)/(x^2-1) = (1+1/x^2)/(1-1/x^2)

Then note that 1/x^2 -> 0 as x->+-oo

So

(x^2+1)/(x^2-1) -> (1+0)/(1-0) = 1 as x->+-oo

So the horizontal asymptote is y = 1

Alternatively, separate out the "polynomial part" 1 from the rational expression as follows:

(x^2+1)/(x^2-1) = (x^2-1+2)/(x^2-1) = 1 + 2/(x^2-1)

Then note that 2/(x^2-1)->0 as x->+-oo

So again we see that the horizontal asymptote is y=1