Question

What are the asymptote(s) and hole(s), if any, of `f(x) =x/(x^3-x)`?

Answer 1

Holes 0
Vertical Asymptotes `+-1`
Horizontal Asymptotes 0

Explanation:

An vertical asymptote or a hole is created by a point in which the domain is equal to zero i.e. `x^3-x=0`

`x(x^2-1)=0`

So either `x=0` or `x^2-1=0`

`x^2-1=0` therefore `x=+-1`

An horizontal asymptote is created where the top and the bottom of the fraction don't cancel out. Whilst a hole is when you can cancel out.

So `color(red)x/(color(red)x(x^2-1))=1/(x^2-1)`

So as the `x` crosses out 0 is merely a hole. Whilst as the `x^2-1` remains `+-1` are asymptotes

For horizontal asymptotes one is trying to find what happens as x approaches infinity or negative infinity and whether it tends to a particular y value.

To do this divide both the numerator and denominator of the fraction by the highest power of `x` in the denominator

`limxtooo(x/(x^3))/(x^3/x^3-x/x^3)=limxtooo(1/(x^2))/(1-1/x^2)=(1/(oo^2))/(1-1/oo^2)=0/(1-0)=0/1=0`

To do this we have to know two rules

`limxtooox^2=oo`
and
`limxtooo1/x^n=1/oo=0 if n>0`

For limits to negative infinty we have to make all the `x` into `-x`

`limxtooo=-x/(-x^3+x)=(-x/(x^3))/(-x^3/x^3+x/x^3)=limxtooo(-1/(x^2))/(-1+1/x^2)=(-1/(oo^2))/(-1+1/oo^2)=0/(-1+0)=0/-1=0`

So the horizontal asymptote as x approaches `+-oo` is 0