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: 28


($\frac{-3}{m}$+3, $\frac{3}{m}$+3), 𝛿 =$\frac{3}{m}$

Work Step by Step

Solve the inequality |f(x) -L| < e |mx-3m| <3 |m(x-3)| <3 -3< 0; Now we need to find the value of 𝛿>0 that places the open interval (3-𝛿, 3+𝛿) centered at 3 inside the interval($\frac{-3}{m}$+3, $\frac{3}{m}$+3). distance between 3 and $\frac{-3}{m}$+3 is$\frac{3}{m}$ and distance between 3 and $\frac{3}{m}$+3 is $\frac{3}{m}$. so we will take the smaller value for 𝛿, so 𝛿 =$\frac{3}{m}$
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