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: 23



Work Step by Step

1) Since $2x-1\leq f(x)\leq x^2$ for $0\lt x\lt3$ and $1\in(0,3)$ 2) Consider $\lim\limits_{x\to1}(2x-1)=2\times 1-1=1$ $\lim\limits_{x\to1}x^2=1^2=1$ That means $\lim\limits_{x\to1}(2x-1)=\lim\limits_{x\to1}x^2=1$ 3) From 1) and 2), according to the Squeeze Theorem, we can conclude that $$\lim\limits_{x\to1}f(x)=1$$
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