Answer
$x=-1$ is a removable discontinuity for the function $f(x)$.
Work Step by Step
Since$$f(x)=\frac{x^2-1}{x+1}=\frac{(x-1)(x+1)}{x+1}=x-1 \qquad x \neq -1,$$ the function $f(x)$ has a removable discontinuity at $x=-1$.
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 1 - Limits and Their Properties - 1.5 Exercises - Page 88: 29
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