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: 43


Discontinuous at $ x=0$

Work Step by Step

Recall that if $ f $ is a polynomial function, then we have $\lim_\limits{x\to a}f(x)=f(a)$. $\lim_\limits{x \to 0^{-}} f(x)= \lim_\limits{x \to 0^{+}}\dfrac{x^2+5x}{x^2-5x}=-1$ and $\lim_\limits{x \to 0^{+}} f(x)= \lim_\limits{x \to 0^{+}}\dfrac{x^2+5x}{x^2-5x}=-1$ So, $\lim_\limits{x \to 0^{-}} f(x)= \lim_\limits{x \to 0^{+}}f(x)$ exists. Now, $\lim_\limits{x \to 0^{-}} f(x)=-1$ and $ f(0)=-2$ so, $\lim_\limits{x \to 0} f(x) \neq f(0)$ Therefore, the function is Discontinuous at $ x=0$
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