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 100: 2

Answer

See graph and explanations below.
1575125953

Work Step by Step

Step 1. See graph for the piecewise function: $f(x)=\begin{cases} 0\hspace1,3cm x\leq-1 \\ 1/x\hspace0.8cm 0\lt |x|\lt1 \\ 0\hspace1.3cm x=1 \\1\hspace1.3cm x\gt1 \\ \end{cases}$ Step 2. For $x=-1$, $\lim_{x\to-1^-}f(x)=0$, $\lim_{x\to-1^+}f(x)=(1/(-1))=-1\ne \lim_{x\to-1^-}f(x)$, thus $\lim_{x\to-1}f(x)$ does not exist, and the function is not continuous at this point. Step 3. For $x=0$, $\lim_{x\to^-}f(x)=-\infty$, $\lim_{x\to0^+}f(x)=\infty$, thus $\lim_{x\to0}f(x)$ does not exit, and the function is not continuous at this point. Step 4. For $x=1$, $\lim_{x\to1^-}f(x)=(1/1)=1$, $\lim_{x\to1^+}f(x)=1$, thus $\lim_{x\to-1}f(x)=1$. However, the function is not continuous at this point as $f(1)=0$, but the discontinuity can be removed if we redefine $f(1)=1$ at this point.
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