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: 32

Answer

L=1, 𝛿 = 0.01

Work Step by Step

L=$\lim\limits_{x \to -1}$ -3x-2 = -3.(-1) -2= 3 - 2= 1 Solve the inequality |f(x) -L| < e |-3x-2-1| <0.03 |-3x-3| <0.03 -0.030; Now we need to find the value of 𝛿>0 that places the open interval (-1-𝛿, -1+𝛿) centered at -1 inside the interval (-1.01, -0.99). distance between -1 and -1.01 is 0.01 and distance between -1 and -0.99 is 0.01. so we will take the smaller value for 𝛿, so 𝛿 = 0.01
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