How do you identify all of the asymptotes of G(x) = (x-1)/(x-x^3)G(x)=x−1x−x3?

1 Answer
George C.
Sep 13, 2015

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

Explanation:

G(x) = (x-1)/(x-x^3) = -(x-1)/(x(x-1)(x+1)) = -1/(x(x+1))G(x)=x−1x−x3=−x−1x(x−1)(x+1)=−1x(x+1)

with exclusion x != 1x≠1

When x = 0x=0 or x = -1x=−1, the denominator is zero and the numerator is non-zero, so these are vertical asymptotes.

G(1) = 0/0G(1)=00 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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