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

2 Answers
Jul 21, 2017

Answer:

The line #x=0# is a vertical asymptode of #f#

The line #x=-1# is a vertical asymptode of #f#

So the line #y=0# is the horizontal asymptode of #f# in both #+oo# and #-oo#

Explanation:

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

Now let's find first the vertical asymtodes of #f# :

#lim_{xrarr1}f(x)=lim_{xrarr1}(x-1)/(x(1-x)(1+x))=#

#lim_{xrarr1}-cancel(1-x)/(xcancel((1-x))(1+x))=lim_{xrarr1}-1/(x(1+x))#

#-1/(1*2)=-1/2#

Because #-1/2inRR# there is no vertical asymtode at #x=1#

#lim_{xrarr0^+}f(x)=lim_{xrarr0^+}-1/(x(1+x))=-oo#

So the line #x=0# is a vertical asymptode of #f#

#lim_{xrarr-1^+}f(x)=lim_{xrarr-1^+}-1/(x(1+x))=-oo#

Also the line #x=-1# is a vertical asymptode of #f#

Now for the oblique and horizontal ones :

If #lim_{xrarrpmoo}f(x)/x=m inRR# and #lim_{xrarrpmoo}(f(x)-mx)=b inRR#

Then the line #y=mx+b# is an oblique asyptode at #pmoo# respectivly

If #m=0# then the asymptode is horizontal

#lim_{xrarrpmoo}f(x)/x=lim_{xrarrpmoo}(x-1)/(x^2-x^4)=lim_{xrarrpmoo}x/(-x^4)=0#

#lim_{xrarrpmoo}f(x)=lim_{xrarrpmoo}(x-1)/(x-x^3)=0#

So the line #y=0# is the horizontal asymptode of #f# in both #+oo# and #-oo#

Jul 21, 2017

Answer:

#"vertical asymptote at " x=0" and " x=-1#
#"horizontal asymptote at "y=0#

Explanation:

#"simplifying f(x)"#

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

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

#"the removal of the factor "(1-x)" from the "#
#"numerator/denominator indicates a hole at "x=1#

The denominator of f(x) cannot be zero as this would make f(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(1+x)=0rArrx=0" and "x=-1#

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

#"horizontal asymptotes occur as"#

#lim_(xto+-oo),f(x)toc" ( a constant)"#

divide terms on numerator/denominator by the highest power of x. that is #x^2#

#f(x)=-(1/x^2)/(x/x^2+x^2/x^2)=-(1/x^2)/(1/x+1)#

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

#rArry=0" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no oblique asymptotes.
graph{-(1)/(x(x+1) [-10, 10, -5, 5]}