University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises - Page 67: 64



Work Step by Step

- Calculate $\lim_{x\to0}(2-x^2)$ and $\lim_{x\to0}(2\cos x)$ $\lim_{x\to0}(2-x^2)=2-0^2=2$ $\lim_{x\to0}(2\cos x)=2\times\cos0=2\times1=2$ So, $\lim_{x\to0}(2-x^2)=\lim_{x\to0}(2\cos x)=2$ - Yet $2-x^2\le g(x)\le 2\cos x$ for all $x$ Therefore, applying the Sandwich Theorem, we conclude that $$\lim_{x\to0}g(x)=2$$
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