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: 51

Answer

No discontinuities.

Work Step by Step

For this piece-wise function, the only possible point of discontinuity is $x=1\to$ $\lim\limits_{x\to1^+}f(x)=(1^+)^2=1;\lim\limits_{x\to1^-}f(x)=(1^-)=1.$ Since $\lim\limits_{x\to1^+}f(x)=;\lim\limits_{x\to1^-}f(x)\to;\lim\limits_{x\to1}f(x)=1.$ Furthermore the limit is equal to the value of the function at that point hence the function is continuous at $x=1$ and does not have any discontinuities at all.
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