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

1 Answer
George C.
Sep 13, 2015

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

Explanation:

#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.

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