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: 57

Answer

$f(x)$ is not continuous at $1$.

Work Step by Step

When $f(1)=\lim\limits_{x\to 1}f(x)$, then $f(x)$ will be continuous at $x=1$. $\lim\limits_{x\to 1^+}f(x)=\lim\limits_{x\to 1^+}\dfrac{x^3-1}{x^2-1} \\=\lim\limits_{x\to 1^+}\dfrac{(x-1)(x^2+x+1)}{(x-1)(x+1)}\\=\lim\limits_{x\to 1^+}\dfrac{3}{x+1}\\=\lim\limits_{x\to 1^+}\dfrac{3}{1+1}\\=\dfrac{3}{2}$ Since, $f(1)=2$, we can see that our result does not satisfy the statement because $\dfrac{3}{2}\ne2$. Therefore, $f(x)$ is not continuous at $1$.
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