Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 13 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Section 13.3 One-sided Limits; Continuous Functions - 13.3 Assess Your Understanding - Page 910: 58


$f(x)$ is not continuous at $2$.

Work Step by Step

When $f(2)=\lim\limits_{x\to 2}f(x)$, then $f(x)$ will be continuous at $x=2$ . $\lim\limits_{x\to 2^-}f(x)=\lim\limits_{x\to 2^-}\dfrac{x^2-2x}{x-2} \\=\lim\limits_{x\to 2^-}\dfrac{x(x-2)}{(x-2)}\\=2$ and $\lim\limits_{x\to 2^+}f(x)=\lim\limits_{x\to 2^+}\dfrac{x-4}{x-1} =-2$ And $f(2)=2$ We can see that our result does not satisfy the statement because $-2 \ne 2$. Therefore, $f(x)$ is not continuous at $2$.
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