Question

How do you find vertical, horizontal and oblique asymptotes for `( 4x^5)/(x^3-1)`?

Answer 1

This function has a vertical asymptote `x=1` and no horizontal or oblique asymptotes.

Explanation:

Given:

`f(x) = (4x^5)/(x^3-1)`

Note that:

`(4x^5)/(x^3-1)=(4x^5-4x^2+4x^2)/(x^3-1)`

`color(white)((4x^5)/(x^3-1))=(4x^2(x^3-1)+4x^2)/(x^3-1)`

`color(white)((4x^5)/(x^3-1))=4x^2+(4x^2)/(x^3-1)`

`color(white)((4x^5)/(x^3-1))=4x^2+(4x^2)/((x-1)(x^2+x+1))`

The only real zero of the denominator is `x=1`, which is not a zero of the numerator.

Hence, `f(x)` has a vertical asymptote at `x=1`

Note also that:

`lim_(x->+-oo) (4x^2)/(x^3-1) = lim_(x->+-oo) (4/x)/(1-1/x^3) = 0/(1-0) = 0`

Hence `f(x)` is asymptotic to the parabola `4x^2` and has no horizontal or oblique asymptotes.

graph{(4x^5)/(x^3-1) [-5, 5, -35, 75]}