Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Review - Exercises - Page 167: 24



Work Step by Step

$$A=\lim\limits_{x\to0}x^2\cos(1/x^2)$$ 1) We know that $$-1\le\cos(1/x^2)\le1$$ then $$-x^2\le x^2\cos(1/x^2)\le x^2$$ (since $x^2\ge0$, the inequality direction does not change) 2) Consider $\lim\limits_{x\to0}(-x^2)=-0^2=0$ and $\lim\limits_{x\to0}(x^2)=0^2=0$ So $\lim\limits_{x\to0}(-x^2)=\lim\limits_{x\to0}(x^2)=0$ 3) From 1) and 2), according to the Squeeze Theorem, we can conclude that $$\lim\limits_{x\to0}x^2\cos(1/x^2)=0$$
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