How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)=(x^3-x)/(x^3-4x)#?

1 Answer
Nov 4, 2016

Answer:

The vertical asymptotes are #x=2# and #x=-2#
The horizontal asymptote is #y=1#

Explanation:

Let's do some simplification
#f(x)=(x^3-x)/(x^3-4x)=(x(x^2-1))/(x(x^2-4))=((x+1)(x-1))/((x+2)(x-2))#

As we cannot divide by #0#,

The degree of the numerator is identical to the degree of the denominator, so there are no oblique asymptotes.
Limit #f(x)=x^2/x^2=1#
#x->+-oo#

So a horizontal asymptote is #y=1#
graph{(x^2-1)/(x^2-4) [-10, 10, -5, 5]}