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: 26

Answer

(20, 30) and 𝛿 =4

Work Step by Step

Solve the inequality |f(x) -L| < e |$\frac{120}{x}$ -5| <1 -1 0; Now we need to find the value of 𝛿>0 that places the open interval (24-𝛿, 24+𝛿) centered at 24 inside the interval (20,30). distance between 24 and 20 is 4 and distance between 24 and 30 is 6. so we will take the smaller value for 𝛿, so 𝛿 =4
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