How do you find the asymptote(s) or holes of # f(x) = (x-1) / (x^2-1)#?

1 Answer
May 18, 2016

Answer:

We have a vertical asymptote at #x+1=0# and no horizontal or slanting asymptote. We have a hole at #x=1#.

Explanation:

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

Although #1/(x+1)# may be defined for #x=1#, as #x-1# has cancelled out, the fact is for #x=1#, we do not have #(x-1)/(x^2-1)#. Hence, we have a hole a #x=1#.

As we have #(x+1)# in denominator, we have a vertical asymptote #x+1=0#.

As the degree of numerator is less than that of denominator, we do not have any horizontal or slanting asymptote.

graph{(x-1)/(x^2-1) [-10, 10, -5, 5]}