Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Review - Exercises: 53


$$\lim\limits_{x\to a}f(x)=0$$

Work Step by Step

For all x, we have $$|f(x)|\le g(x)$$ which means $$-g(x)\le f(x)\le g(x)\hspace{1cm}(1)$$ Also, we have $\lim\limits_{x\to a}-g(x)=-\lim\limits_{x\to a}g(x)=-0=0=\lim\limits_{x\to a}g(x)\hspace{1cm}(2)$ Therefore, from (1) and (2), according to Squeeze Theorem, we can conclude that $$\lim\limits_{x\to a}f(x)=0$$
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