# What are the asymptotes for #G(x) = (x-1)/(x-x^3)#?

##### 2 Answers

You may start by factorizing the denominator:

#### Explanation:

So **vertical** asympotes.

If we realize that

And this goes to zero if

So **horizontal** asymptote.

graph{(x-1)/(x-x^3) [-16.6, 15.44, -7.7, 8.33]}

#### Explanation:

#"simplifying " g(x)#

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

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

#"the removal of " (x-1)" from the numerator/denominator"#

#"means there is a hole at x = 1"#

#"the graph of " -1/(x(x+1))" is the same as " (x-1)/(x-x^3)#

#"but without the hole"# The denominator of g(x) cannot be zero as this would make g(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(x+1)=0rArrx=0,x=-1#

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

#"Horizontal asymptotes occur as"#

#lim_(xto+-oo),g(x)toc" ( a constant)"# divide terms on numerator/denominator by the highest power of x, that is

#x^2#

#g(x)=-(1/x^2)/(x^2/x^2+x/x^2)=-(1/x^2)/(1+1/x)# as

#xto+-oo,g(x)to0/(1+0)#

#rArry=0" is the asymptote"#

graph{(x-1)/(x-x^3) [-10, 10, -5, 5]}