Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.4 - One-Sided Limits - Exercises 2.4 - Page 75: 49

Answer

See explanations.

Work Step by Step

Step 1. For a given small value $\epsilon\gt0$, we need to find a value $\delta\gt0$ so that for all $x$ in the interval of $-\delta\lt x\lt0$, we get $|\frac{x}{|x|}+1|\lt\epsilon$ Step 2. As $x\lt0$, we have $|x|=-x$; the last inequality becomes $|-\frac{x}{x}+1|\lt\epsilon$, which gives $0\lt\epsilon$ Step 3. As the result from step 2 is always true, we can set $\delta$ to be any value such as $\delta=\epsilon$ so that for all $x$ in the interval of $-\delta\lt x\lt0$, we get $|\frac{x}{|x|}+1|\lt\epsilon$, which proves the limit statement.
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