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: 27


($\frac{-0.03}{m}$+2, $\frac{0.03}{m}$+2) 𝛿 =$\frac{0.03}{m}$

Work Step by Step

Solve the inequality |f(x) -L| < e |mx-2m| <0.03 |m(x-2)| <0.03 -0.03< 0; Now we need to find the value of 𝛿>0 that places the open interval (2-𝛿, 2+𝛿) centered at 2 inside the interval ($\frac{-0.03}{m}$+2, $\frac{0.03}{m}$+2). distance between 2 and $\frac{-0.03}{m}$+2 is $\frac{0.03}{m}$ and distance between 2 and $\frac{0.03}{m}$ is $\frac{0.03}{m}$. so we will take the smaller value for 𝛿, so 𝛿 =$\frac{0.03}{m}$
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.