## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 2 - Section 2.3 - Calculating Limits Using the Limit Laws - 2.3 Exercises: 41

#### Answer

$\lim\limits_{x\to3}(2x+|x-3|)=6$

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

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