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

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Answer 1 · 2016-11-17T07:20:18

The vertical asymptotes are $x=0$ and $x=-1$

The horizontal asymptote is $y=0$

Let's simplify the expression

$(x-1)/(x-x^3)=(x-1)/(x(1-x^2))=(x-1)/(x(1+x)(1-x))$

$=-cancel(1-x)/(x(1+x)cancel(1-x))=-1/(x(1+x))$

So, we have a hole at $(x=1)$

A we cannot divide by $0$ ,

$x!=0$ and $x!=-1$

We have two vertical asymptotes $x=0$ and $x=-1$

As the degree of the numerator $<$ degree of the denominator,
we don't have oblique asymptotes.

$lim_(x->+-oo)-1/x^2=0^(-)$

The horizontal asymptote is $y=0$
graph{-1/(x(1+x)) [-25.66, 25.65, -12.83, 12.84]}

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