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: 25

Answer

($\sqrt 15$, $\sqrt 17$) and 𝛿 = 0.1231

Work Step by Step

Solve the inequality |f(x) -L| < e |$x^{2}$ - 5-11| <1 -1 x > $\sqrt 17$ or $\sqrt 15$< - x < $\sqrt 17$ -$\sqrt 15$< x < -$\sqrt 17$ Since we need interval about c=4, required interval is($\sqrt 15$, $\sqrt 17$) Find a value of 𝛿>0; Now we need to find the value of 𝛿>0 that places the open interval (4-𝛿, 4+𝛿) centered at 4 inside the interval ($\sqrt 15$, $\sqrt 17$). distance between 4and $\sqrt 15$ is 0.12702 and distance between 4 and $\sqrt 17$ is]0.1231. so we will take the smaller value for 𝛿, so 𝛿 = 0.1231
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