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 - Chapter Review - Review Exercises - Page 925: 13

Answer

The function $f(x)$ is not continuous at $c=-2$.

Work Step by Step

We are given: $f(x)=\dfrac{x^2-4}{x+2}$ We need to determine if $f(x)$ is continuous at $c=-2$. We check the left-hand and right-hand limits. If they are equal to each other and the value of the function at the limit, then the function is continuous at that point. $\lim\limits_{x \to 2^{-}}f(x)=\lim\limits_{x \to 2^{-}}\dfrac{x^2-4}{x+2} \\=\dfrac{(-2)^2-4}{-2+2} \\=\dfrac{0}{0}$ We got an undefined value, so the function cannot be defined at $c=-2$. Therefore, the function $f(x)$ is not continuous at $c=-2$.
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