What are the asymptote(s) and hole(s), if any, of # f(x) =x/(x^3-x)#?

1 Answer
Mar 30, 2018

Answer:

Holes 0
Vertical Asymptotes #+-1#
Horizontal Asymptotes 0

Explanation:

An vertical asymptote or a hole is created by a point in which the domain is equal to zero i.e. #x^3-x=0#

#x(x^2-1)=0#

So either #x=0# or #x^2-1=0#

#x^2-1=0# therefore #x=+-1#

An horizontal asymptote is created where the top and the bottom of the fraction don't cancel out. Whilst a hole is when you can cancel out.

So #color(red)x/(color(red)x(x^2-1))=1/(x^2-1)#

So as the #x# crosses out 0 is merely a hole. Whilst as the #x^2-1# remains #+-1# are asymptotes

For horizontal asymptotes one is trying to find what happens as x approaches infinity or negative infinity and whether it tends to a particular y value.

To do this divide both the numerator and denominator of the fraction by the highest power of #x# in the denominator

#limxtooo(x/(x^3))/(x^3/x^3-x/x^3)=limxtooo(1/(x^2))/(1-1/x^2)=(1/(oo^2))/(1-1/oo^2)=0/(1-0)=0/1=0#

To do this we have to know two rules

#limxtooox^2=oo#
and
#limxtooo1/x^n=1/oo=0 if n>0#

For limits to negative infinty we have to make all the #x# into #-x#

#limxtooo=-x/(-x^3+x)=(-x/(x^3))/(-x^3/x^3+x/x^3)=limxtooo(-1/(x^2))/(-1+1/x^2)=(-1/(oo^2))/(-1+1/oo^2)=0/(-1+0)=0/-1=0#

So the horizontal asymptote as x approaches #+-oo# is 0