Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.2 Exercises - Page 58: 64

Answer

$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline x & -0.1 & -0.01 & - 0.001 & 0 & 0.001 & 0.01 & 0.1 \\ \hline f(x) & 2 & 2 & 2 & \text{undefined} & 2 & 2 & 2 \\ \hline \end{array}$$

Work Step by Step

Looking at the graph and table, we find that as $x$ approaches $0$ from the left and right, th function $f(x)$ approaches $2$. Thus, we can conclude that$$\lim_{x \to 0}\frac{|x+1|-|x-1|}{x}=2 .$$
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