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

Question

763 views around the world

You can reuse this answer
Creative Commons License

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]}