How do you find the asymptotes for #G(x) = (x-1)/(x-x^3)#?

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Answer 1 · 2016-11-23T12:09:25

There is a hole at #x=-1#
The vertical asymptotes are #x=0# and #x=1#
No oblique asymptote.
The horizontal asymptote is #y=0#

Let's factorise the numerator and denominator

#x-1=-(1+x)#

#x-x^3=x(1-x^2)=x(1-x)(1+x)#

So,

#G(x)=(x-1)/(x(1-x^2))=-cancel(1+x)/(x(1-x)cancel(1+x))#

#=-1/(x(1-x))#

There is a hole at #x=-1#

As we cannot divide by #0#,
so, #x!=0# and #x!=1#

Therefore, #x=0# and #x=1# are vertical asymptotes

As the degree of the numerator is #># the degree of the denominator, there is no slant asymptote.

For the limits, we take the terms of highest degree-

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

Therefore, #y=0# is a horizontal asymptote

graph{(x-1)/(x(1-x^2)) [-18.16, 13.88, -7.75, 8.27]}