How do you identify all of the asymptotes of #G(x) = (x-1)/(x-x^3)#?

Question

Benjamin Achtenberg · Accepted Answer

Find #G(x) = -1/(x(x+1))# with exclusion #x != 1#, hence vertical asymptotes #x = -1# and #x = 0# and horizontal asymptote #y = 0#.

#G(x) = (x-1)/(x-x^3) = -(x-1)/(x(x-1)(x+1)) = -1/(x(x+1))# with exclusion #x != 1# When #x = 0# or #x = -1#, the denominator is zero and the numerator is non-zero, so these are vertical asymptotes. #G(1) = 0/0# is undefined. This is a removable singularity - not an asymptote. #lim_(x->1) G(x) = -1/2# exists. As #x->oo, G(x)->0#, so #y = 0# is a horizontal asymptote.

Theodore Amidon · Answer

undefined To identify the asymptotes of $ G(x) = \frac{x - 1}{x - x^3} $, follow these steps:

1. Vertical Asymptotes:
   - Set the denominator equal to zero and solve for $ x $.
   - If any solution is not a removable singularity (i.e., the factor does not cancel out in both the numerator and the denominator), then it represents a vertical asymptote.

2. Horizontal Asymptotes:
   - Compare the degrees of the numerator and the denominator.
   - If the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at $ y = 0 $.
   - If the degree of the numerator equals the degree of the denominator, divide the leading coefficients to find the horizontal asymptote.

3. Oblique (Slant) Asymptotes:
   - If the degree of the numerator is exactly one more than the degree of the denominator, there is an oblique asymptote.
   - To find the equation of the oblique asymptote, perform polynomial long division or use synthetic division to divide the numerator by the denominator, and the quotient will represent the equation of the oblique asymptote.

Following these steps will help you identify all the asymptotes of the given rational function.