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Chop Suey (x-1)/(x (1-x^2) = (x-1)/(x (1-x)(1+x) ) = -1/(x (1+x) )
Take note that since we canceled 1-x, then there is a hole at x = 1. That is, f(x) is not defined at x = 1.
There is no oblique asymptotes.
The vertical asymptote is at x = 0 and x = -1 (These are the points that makes the function undefined).
The horizontal asymptote is zero since the limit as x goes to ± infinity equals 0. In Algebra, you would probably know it by this way:
- Degree of Numerator > Degree of Denominator, then no horizontal asymptote exist.
- Degree of Numerator = Degree of Denominator, then horizontal asymptotes is the y = a/b, where a and b is the coefficients for the leading terms in numerator and denominator, respectively.
- Degree of Numerator < Degree of Denominator, then horizontal asymptote is y = 0