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: 61

Answer

$f(x)$ is continuous over the interval $(-\infty, 1)$ U $(1, \infty).$

Work Step by Step

The only possible point of discontinuity is $x=1\to$ $\lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^-}\dfrac{3x^2-x-2}{x-1}=\lim\limits_{x\to1^-}(3x+2)=3(1^-)+2=5.$ $\lim\limits_{x\to1^+}f(x)=\lim\limits_{x\to1^+}\dfrac{3x^2-x-2}{x-1}=\lim\limits_{x\to1^+}(3x+2)=3(1^+)+2=5.$ Since $\lim\limits_{x\to1^+}f(x)=\lim\limits_{x\to1^-}f(x)\to\lim\limits_{x\to1}f(x)$ exists and is equal to $5.$ But since $\lim\limits_{x\to1}f(x)\ne f(1)$, the function has a discontinuity at $x=1.$ The function is therefore continuous over the interval $(-\infty, 1)$ U $(1, \infty).$
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