What are the holes of f(x)= (x^3-x)/(4x^3-16x)?

Are there any holes , if yes how do you get them?
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x(x^2-1)/(4x(x^2-4))=x(x-1)(x+1)/(4x(x-2)(x+2)). This is (x-1)(x+1)/(4(x-2)(x+2)) if x not zero.

Hole at x=0 because of common x in numerator and denominator. The expression approaches 1/16.

(At x=2 or -2 there are asymptotes.)

by Top Rated User (1.1m points)

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