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 - 1.4 Exercises: 48

Answer

$f(x)$ has a removable discontinuity at $x=-2$ and a non-removable discontinuity at $x=3.$

Work Step by Step

$f(x)=\dfrac{x+2}{x^2-x-6}=\dfrac{(x+2)}{(x+2)(x-3)}=\dfrac{1}{x-3}\to x\ne-2$ and $x\ne3.$ Hence, $f(x)$ has a removable discontinuity at $x=-2$ and an irremovable discontinuity (vertical asymptote) at $x=3.$
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