Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.3 - Calculating Limits Using the Limit Laws - 2.3 Exercises - Page 103: 37



Work Step by Step

We have $\lim\limits_{x\to4}(4x-9)=4\times4-9=7$ $\lim\limits_{x\to4}(x^2-4x+7)=4^2-4\times4+7=7$ Therefore, $\lim\limits_{x\to4}(4x-9)=\lim\limits_{x\to4}(x^2-4x+7)=7$ But, $4x-9\leq f(x)\leq x^2-4x+7$ for $x\geq0$ So, applying the squeeze theorem, we have $\lim\limits_{x\to4}f(x)=7$
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