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: 56

Answer

The function does not have any discontinuities.

Work Step by Step

$|x-3|\leq2\to-2\leq x-3\leq2\to1\leq x\leq5\to$ the possible x-values for discontinuities are $x=1$ and $x=5:$ $\lim\limits_{x\to1^-}f(x)=2; \lim\limits_{x\to1^+}f(x)=\csc{\dfrac{\pi (1^+)}{6}}=2.$ Since $\lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^+}f(x)\to\lim\limits_{x\to1}f(x)=2=f(1)$ hence the function is continuous at $x=1.$ $\lim\limits_{x\to5^-}f(x)=\csc{\dfrac{\pi(5^-)}{6}}=2; \lim\limits_{x\to5^+}f(x)=2.$ Since $\lim\limits_{x\to5^-}f(x)=\lim\limits_{x\to5^+}f(x)\to\lim\limits_{x\to5}f(x)=2=f(5)$ hence the function is continuous at $x=5.$ As the function is continuous at both points, it does not have any discontinuities.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.