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 - Page 80: 46

Answer

A removable discontinuity at $x=5$ and a non-removable discontinuity at $x=-5$.

Work Step by Step

$f(x)=\dfrac{(x-5)}{(x-5)(x+5)}=\dfrac{1}{x+5}\to x\ne5$ and $x\ne-5$. Hence, we have an non-removable discontinuity at $x=-5$ and a removable discontinuity at $x=5.$ A graph of $\frac{1}{x+5}$ will make the situation clear.
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