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

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Answer 1 · 2016-01-07T19:26:17

See 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#