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

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

Now let's find first the vertical asymtodes of `f` :

`lim_{xrarr1}f(x)=lim_{xrarr1}(x-1)/(x(1-x)(1+x))=`

`lim_{xrarr1}-cancel(1-x)/(xcancel((1-x))(1+x))=lim_{xrarr1}-1/(x(1+x))`

`-1/(1*2)=-1/2`

Because `-1/2inRR` there is no vertical asymtode at `x=1`

`lim_{xrarr0^+}f(x)=lim_{xrarr0^+}-1/(x(1+x))=-oo`

So the line `x=0` is a vertical asymptode of `f`

`lim_{xrarr-1^+}f(x)=lim_{xrarr-1^+}-1/(x(1+x))=-oo`

Also the line `x=-1` is a vertical asymptode of `f`

Now for the oblique and horizontal ones :

If `lim_{xrarrpmoo}f(x)/x=m inRR` and `lim_{xrarrpmoo}(f(x)-mx)=b inRR`

Then the line `y=mx+b` is an oblique asyptode at `pmoo` respectivly

If `m=0` then the asymptode is horizontal

`lim_{xrarrpmoo}f(x)/x=lim_{xrarrpmoo}(x-1)/(x^2-x^4)=lim_{xrarrpmoo}x/(-x^4)=0`

`lim_{xrarrpmoo}f(x)=lim_{xrarrpmoo}(x-1)/(x-x^3)=0`

So the line `y=0` is the horizontal asymptode of `f` in both `+oo` and `-oo`

`"simplifying f(x)"`

`f(x)=(x-1)/(x(1-x^2))=-(cancel((1-x)))/(xcancel((1-x))(1+x))`

`rArrf(x)=-1/(x(1+x))`

`"the removal of the factor "(1-x)" from the "`
`"numerator/denominator indicates a hole at "x=1`

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

`"solve "x(1+x)=0rArrx=0" and "x=-1`

`rArrx=0" and " x=-1" are the asymptotes"`

`"horizontal asymptotes occur as"`

`lim_(xto+-oo),f(x)toc" ( a constant)"`

divide terms on numerator/denominator by the highest power of x. that is `x^2`

`f(x)=-(1/x^2)/(x/x^2+x^2/x^2)=-(1/x^2)/(1/x+1)`

as `xto+-oo,f(x)to0/(0+1)`

`rArry=0" is the asymptote"`

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no oblique asymptotes.
graph{-(1)/(x(x+1) [-10, 10, -5, 5]}