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 2.3 - The Precise Definition of a Limit - Exercises 2.3 - Page 66: 47


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Work Step by Step

Step 1. We can identify the quantities as: $c=1,L=2$. $f_1(x)=4-2x$ when $x\lt1$, and $f_2(x)=6x-4$ when $x\geq1$. Step 2. For a given small value $\epsilon\gt0$, we need to find a value $\delta\gt0$ so that for all $x$ when $0\lt |x-1|\lt\delta$, we have $|f(x)-2|\lt\epsilon$ which gives either $|2-2x|\lt\epsilon$ or $|6x-6|\lt\epsilon$ Step 3. The last two inequality gives $-\epsilon\lt 2-2x \lt\epsilon$ or $1-\frac{\epsilon}{2}\lt x \lt 1+\frac{\epsilon}{2}$. Also, $-\epsilon\lt 6x-6 \lt\epsilon$ or $1-\frac{\epsilon}{6}\lt x \lt 1+\frac{\epsilon}{6}$. These inequalities can be combined as $1-\frac{\epsilon}{6}\lt x \lt 1+\frac{\epsilon}{6}$ Step 4. From the requirement for the value $\delta$ in step 2, we have $-\delta\lt x-1\lt\delta$ or $1-\delta\lt x\lt1+\delta$ Step 5. Compare the results from steps 4 and 3 ,we can set $\delta=\frac{1}{6}\epsilon$ Step 6. With the setting of the $\delta$ value, we can go backwards through the above steps to reach a conclusion that for all $x$ when $0\lt |x-1|\lt\delta$, we have $|f(x)-2|\lt\epsilon$ which proves the limit statement.
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