Precalculus (6th Edition) Blitzer

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

Chapter 11 - Test - Page 1180: 11


$ f $ is discontinuous at $x=-1$

Work Step by Step

Recall that if $ f $ is a polynomial function, then we have $\lim_\limits{x\to a}f(x)=f(a)$. $\lim_\limits{x\to -1} f(x)=\lim_\limits{x\to -1} \dfrac{x^2-1}{x+1}$ $\lim_\limits{x\to -1} f(x)=\lim_\limits{x\to -1} \dfrac{(x-1)(x+1)}{x+1}=\lim_\limits{x\to -1} (x-1)$ $\lim_\limits{x\to -1} f(x)=(-1) -(1)=-2$ and $ f(-1) =6$ So, $\lim_\limits{x\to -1} f(x) \neq f(-1)$ Therefore, the function $ f $ is discontinuous at $x=-1$
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