Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Practice Exercises - Page 101: 31


No, see explanations.

Work Step by Step

Step 1. For $x\lt-1, x\ne-1$, $f(x)=\frac{x(x^2-1)}{|x^2-1|}==\frac{x(x^2-1)}{x^2-1}=x$, $\lim_{x\to-1^-}f(x)=-1$ Step 2. For $-1\lt x\lt1, x\ne-1, x\ne1$, $f(x)=\frac{x(x^2-1)}{|x^2-1|}==-\frac{x(x^2-1)}{x^2-1}=-x$, $\lim_{x\to-1^+}f(x)=-(-1)=1$, and $\lim_{x\to1^-}f(x)=-(1)=-1$ Step 3. For $x\gt1, x\ne1$, $f(x)=\frac{x(x^2-1)}{|x^2-1|}==\frac{x(x^2-1)}{x^2-1}=x$, $\lim_{x\to1^+}f(x)=1$ Step 4. Graph the function as shown. Step 5. We can see that the continuity can not be extended at $x=1$ or $x=-1$ because the limits do not exist at these points.
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