How do you find vertical, horizontal and oblique asymptotes for #[(e^-x)(x^5) + 2] /[ x^5 - x^4 -x +1]#?

Question

Liam Brennan · Accepted Answer

Define #g(x)=(x^5*e^-x+2)/(x^5-x^4-x+1)# Bottom quintic should be factorised in order to find potential vertical asymptotes.  #f(x)=x^5-x^4-x+1# Can be seen easily that 1 and -1 are factors. Then #(x-1)(x+1)#, #x^2-1# is a factor of #f(x)=0#. Then, #f(x) = (x^2-1)*(Ax^3+Bx^2+Cx+D)#,
#f(x)=Ax^5 +Bx^4 + (C-A)x^3+(D-B)x^2-Cx-D# Comparing coefficients gives, #f(x)=(x^2-1)(x^3-x^2+x-1)#. Another obvious factor of the cubic is -1. Applying the same method, #f(x)=(x-1)^2(x+1)(x^2+1)#. Then we see, #g(x)=(x^5*e^-x+2)/((x-1)^2(x+1)(x^2+1))#. From this, we can deduce there are horizontal asymptotes as the denominator goes to 0.  Vertical asymptotes at #x=-1# and #x=1#. Oblique and horizontal asymptotes arise as #x# goes to #\pm\infty#. Rewrite #g(x)# as, #g(x) = (x^5)/((x-1)^2(x+1)(x^2+1))*e^-x+(2)/((x-1)^2(x+1)(x^2+1))#. Because the numerator and denominator are 5th order polynomials, 
#\lim_{x	o+\infty} ((x^5)/((x-1)^2(x+1)(x^2+1))) = 1#. Due to the nature of exponentials, 
#\lim_{x	o+\infty} e^-x = 0#. Clearly,
#\lim_{x	o+\infty} ((2)/((x-1)^2(x+1)(x^2+1))) = 0#. Then,
#lim_{x	o+\infty}g(x)=1*0+0#
#lim_{x	o+\infty}g(x)=0# This gives us a horizontal asymptote #y=0#. Using the same limits,  #g(x) ightarrow e^-x# as #x ightarrow \infty#.  This gives us an oblique asymptote #y=e^-x#.

Caleb Alexander · Answer

undefined To find the vertical asymptotes, determine the values of $ x $ that make the denominator zero, excluding any values that make the numerator zero as well. For horizontal asymptotes, look at the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. If the degrees are equal, divide the leading coefficients to find the horizontal asymptote. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. To find oblique asymptotes, perform polynomial long division to divide the numerator by the denominator and determine the quotient. The quotient represents the oblique asymptote.