Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 2 - Section 2.4 - The Precise Definition of a Limit - 2.4 Exercises - Page 114: 14

Answer

When $\epsilon = 0.1$, we can let $\delta = 0.02$ When $\epsilon = 0.05$, we can let $\delta = 0.01$ When $\epsilon = 0.01$, we can let $\delta = 0.002$

Work Step by Step

It is given that: $\lim\limits_{x \to 2}(5x-7) = 3$ We can let $\delta = \frac{\epsilon}{5}$ Let $\epsilon = 0.1$ Let $\delta = 0.02$ Suppose that $\vert x-2 \vert \lt \delta$ Then: $\vert (5x-7)-3 \vert = \vert 5x-10 \vert = 5 \vert x-2 \vert \lt 5(\delta) = 5(0.02) = 0.1$ Let $\epsilon = 0.05$ Let $\delta = 0.01$ Suppose that $\vert x-2 \vert \lt \delta$ Then: $\vert (5x-7)-3 \vert = \vert 5x-10 \vert = 5 \vert x-2 \vert \lt 5(\delta) = 5(0.01) = 0.05$ Let $\epsilon = 0.01$ Let $\delta = 0.002$ Suppose that $\vert x-2 \vert \lt \delta$ Then: $\vert (5x-7)-3 \vert = \vert 5x-10 \vert = 5 \vert x-2 \vert \lt 5(\delta) = 5(0.002) = 0.01$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.