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: 44


Continuous at $ x=-1$

Work Step by Step

Recall that if $ f $ is a polynomial function, then we have $\lim_\limits{x\to a}f(x)=f(a)$. $\lim_\limits{x \to -1^{-}} f(x)= \lim_\limits{x \to -1^{-}}\dfrac{x^2+x}{x^2-3x-4}=\dfrac{1}{5}$ and $\lim_\limits{x \to -1^{+}} f(x)= \lim_\limits{x \to -1^{+}}\dfrac{x}{x-4}=\dfrac{1}{5}$ So, $\lim_\limits{x \to -1^{-}} f(x)= \lim_\limits{x \to -1^{+}}f(x)$ exists. Now, $\lim_\limits{x \to-1} f(x)=\dfrac{1}{5}$ and $ f(-1)=\dfrac{1}{5}$ so, $\lim_\limits{x \to 0} f(x) = f(0)$ Therefore, the function is continuous at $ x=-1$
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