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 - Review Exercises: 62

Answer

The function is continuous over the interval $(-\infty, 2)$ U $(2, \infty).$

Work Step by Step

The only possible point of discontinuity is $x=2.$ $\lim\limits_{x\to2^-}f(x)=\lim\limits_{x\to2^-}(5-x)=5-2^-=3.$ $\lim\limits_{x\to2^+}f(x)=\lim\limits_{x\to2^+}(2x-3)=2(2^+)-3=1.$ Since $\lim\limits_{x\to2^-}f(x)\ne\lim\limits_{x\to2^+}f(x)\to\lim\limits_{x\to2}f(x)$ does not exist then the function has a discontinuity at $x=2.$ The function is continuous over the interval $(-\infty, 2)$ U $(2, \infty).$
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