How do you find vertical, horizontal and oblique asymptotes for #( 4x^5)/(x^3-1)#?

1 Answer
Aug 8, 2018

Answer:

This function has a vertical asymptote #x=1# and no horizontal or oblique asymptotes.

Explanation:

Given:

#f(x) = (4x^5)/(x^3-1)#

Note that:

#(4x^5)/(x^3-1)=(4x^5-4x^2+4x^2)/(x^3-1)#

#color(white)((4x^5)/(x^3-1))=(4x^2(x^3-1)+4x^2)/(x^3-1)#

#color(white)((4x^5)/(x^3-1))=4x^2+(4x^2)/(x^3-1)#

#color(white)((4x^5)/(x^3-1))=4x^2+(4x^2)/((x-1)(x^2+x+1))#

The only real zero of the denominator is #x=1#, which is not a zero of the numerator.

Hence, #f(x)# has a vertical asymptote at #x=1#

Note also that:

#lim_(x->+-oo) (4x^2)/(x^3-1) = lim_(x->+-oo) (4/x)/(1-1/x^3) = 0/(1-0) = 0#

Hence #f(x)# is asymptotic to the parabola #4x^2# and has no horizontal or oblique asymptotes.

graph{(4x^5)/(x^3-1) [-5, 5, -35, 75]}