University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.4 - One-Sided Limits - Exercises - Page 85: 47


The interval $I=(5,5+\delta)$ with $\delta=\epsilon^2$ would get $\sqrt{x-5}\lt\epsilon$ The limit being referred to here is $$\lim_{x\to5^+}\sqrt{x-5}=0$$

Work Step by Step

Given $\epsilon\gt0$, first, we look at the inequality: $$\sqrt{x-5}\lt\epsilon$$ - Square both sides: $$x-5\lt\epsilon^2$$ $$x\lt5+\epsilon^2$$ Therefore, if $x\lt5+\epsilon^2$, then $\sqrt{x-5}\lt\epsilon$ Take $\delta=\epsilon^2$, we would have $$x\lt5+\delta\Rightarrow\sqrt{x-5}\lt\epsilon$$ Combining with the interval $I=(5,5+\delta)$ with $\delta=\epsilon^2$: $$5\lt x\lt5+\delta\Rightarrow\sqrt{x-5}\lt\epsilon$$ According to the defintion of one-side limit, the limit being referred to here is $$\lim_{x\to5^+}\sqrt{x-5}=0$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.