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

1 Answer
May 18, 2016

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]}