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: 41



Work Step by Step

$\lim\limits_{x\to3}(2x+|x-3|)$ $=\lim\limits_{x\to3}2x+\lim\limits_{x\to3}|x-3|$ $=6+\lim\limits_{x\to3}|x-3|$ We know that $|x-3|=$$\left \{ \begin{array} {c l} x-3 && x\geq3\\ -(x-3)=3-x && x<3 \end{array} \right.$ Therefore, $\lim\limits_{x\to3^+}|x-3|=\lim\limits_{x\to3^+}(x-3)=3-3=0$ $\lim\limits_{x\to3^-}|x-3|=\lim\limits_{x\to3^-}(3-x)=3-3=0$ We see that $\lim\limits_{x\to3^+}|x-3|=\lim\limits_{x\to3^-}|x-3|=0$ So, $\lim\limits_{x\to3}|x-3|=0$ Therefore, $$\lim\limits_{x\to3}(2x+|x-3|)=6+0=6$$
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