Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.4 - The Precise Definition of a Limit - 2.4 Exercises: 3

Answer

$\delta=1.44$, (or any smaller positive number).

Work Step by Step

Since $1.6^{2}=2.56,\quad (\sqrt{2.56}=1.6\quad f(2.56)=1.6)$ and $2.4^{2}=6.76,\quad (\sqrt{6.76}=2.4\quad f(6.76)=2.4)$ the unknown interval bounds around x=4 are 2.56 and 6.76. Now, from the graph we see that for $x\in(2.56, 6.76)$ or $2.56< x < 6.76$ $-1.44< x-4 < 2.76$ it follows that $1.6 < \sqrt{x} < 2.4$ $-0.4< \sqrt{x} -2 < 0.4$ $|\sqrt{x}-2|<0.4.$ We want a positive number $\delta$ such that $-1.44\leq\ \ -\delta < x-4 < \delta\ \ < 2.76, $ So, if we observe the stricter vicinity of $\pm 1.44 $ around x=4, it follows $-1.44\leq\ \ -1.44 < x-4 < 1.44\ \ < 2.76, $ $|x-4|<1.44$ ensures that x is still in $(2.56, 6.76)$ which also ensures that $|\sqrt{x}-2|<0.4.$ We take $\delta=1.44$, (or any smaller positive number).
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