## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 76: 17

#### Answer

1) The open interval is $(-0.19,0.21)$ 2) $\delta=0.19$

#### Work Step by Step

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-0|\lt\delta\Rightarrow|\sqrt{x+1}-1|\lt0.1$$ 1) Find the interval around $0$ on which $|\sqrt{x+1}-1|\lt0.1$ holds. Solve the inequality: $$|\sqrt{x+1}-1|\lt0.1$$ $$-0.1\lt\sqrt{x+1}-1\lt0.1$$ $$0.9\lt\sqrt{x+1}\lt1.1$$ Square: $$0.81\lt x+1\lt1.21$$ $$-0.19\lt x\lt 0.21$$ The open interval around $0$ is $(-0.19,0.21)$. 2) Give a value for $\delta$ The nearer endpoint to $0$ is $-0.19$, and the distance between them is $0.19$. So if we take $\delta=0.19$ or any smaller positive number, then $0\lt|x-0|\lt0.19$, meaning all $x$ would be placed in the interval $(-0.19,0.21)$ so that $|\sqrt{x+1}-1|\lt0.1$. In other words, $$0\lt |x-0|\lt0.19\Rightarrow|\sqrt{x+1}-1|\lt0.1$$

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