Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - Review Exercises: 53

Answer

$f(x)$ has a removable discontinuity at $x=0$ and two irremovable discontinuities (vertical asymptotes) at $x=1$ and $x=-1.$

Work Step by Step

$f(x)=\dfrac{x}{x^3-x}=\dfrac{1}{x^2-1}=\dfrac{1}{(x-1)(x+1)};x\ne-1,x\ne0,x\ne1.$ At $x=0$ the function has a removable discontinuity (since the factor cancels out from both the numerator and denominator). At both $x=1$ and $x=-1$, the denominator is $0$, indicating the presence of a vertical asymptote at both points. The vertical asymptotes are not removable.
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