Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Review Exercises - Page 1179: 50


Discontinuous at $ x=-2$

Work Step by Step

Recall that if $ f $ is a polynomial function, then we have $\lim_\limits{x\to a}f(x)=f(a)$. $\lim_\limits{x \to -2} f(x)= \lim_\limits{x \to -2} \dfrac{x^2-4}{x+2}= \lim_\limits{x \to -2} \dfrac{(x-2)(x+2)}{x+2}=-4$ and $\lim_\limits{x \to -2} f(x)= \lim_\limits{x \to -2} \dfrac{x^2-4}{x+2}= \lim_\limits{x \to -2} \dfrac{(x-2)(x+2)}{x+2}=-4$ So, $\lim_\limits{x \to 2} f(x)= \lim_\limits{x \to -2}f(x)$ exists. Now, $\lim_\limits{x \to -2} f(x)=-2$ so, $\lim_\limits{x \to -2} f(x) \ne -4$ Therefore, the function is discontinuous at $ x=-2$
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