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

##### 1 Answer

Jan 7, 2016

#### Answer:

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# as#x->+-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