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

Published by Pearson

# Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 76: 24

#### Answer

1) The open interval is $(-10/9,-10/11)$ 2) $\delta=1/11$

#### Work Step by Step

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-(-1)|\lt\delta\Rightarrow|\frac{1}{x}-(-1)|\lt0.1$$ 1) Find the interval around $-1$ on which $|\frac{1}{x}-(-1)|\lt0.1$ holds. Solve the inequality: $$|\frac{1}{x}-(-1)|\lt0.1$$ $$|\frac{1}{x}+1|\lt0.1$$ $$-0.1\lt\frac{1}{x}+1\lt0.1$$ $$-1.1\lt\frac{1}{x}\lt-0.9$$ $$-\frac{1}{1.1}\gt x\gt-\frac{1}{0.9}$$ $$-\frac{10}{11}\gt x\gt-\frac{10}{9}$$ The open interval around $-1$ is $(-10/9,-10/11)$. 2) Give a value for $\delta$ The nearer endpoint to $-1$ is $-10/11$, and the distance between them is $|-1|-|-10/11|=1/11$. So if we take $\delta=1/11$ or any smaller positive number, then $0\lt|x-(-1)|\lt1/11$, meaning all $x$ would be placed in the interval $(-10/11,-10/9)$ so that $|\frac{1}{x}-(-1)|\lt0.1$. In other words, $$0\lt |x-(-1)|\lt1/11\Rightarrow|\frac{1}{x}-(-1)|\lt0.1$$

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