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

Published by Pearson

# Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 76: 26

#### Answer

1) The open interval is $(20,30)$. 2) $\delta=4$

#### Work Step by Step

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-24|\lt\delta\Rightarrow|\frac{120}{x}-5|\lt1$$ 1) Find the interval around $24$ on which $|\frac{120}{x}-5|\lt1$ holds. Solve the inequality: $$|\frac{120}{x}-5|\lt1$$ $$-1\lt\frac{120}{x}-5\lt1$$ $$4\lt\frac{120}{x}\lt6$$ $$\frac{1}{30}\lt \frac{1}{x}\lt\frac{1}{20}$$ $$30\gt x\gt20$$ The open interval around $24$ is $(20,30)$. 2) Give a value for $\delta$ The nearer endpoint to $24$ is $20$, and the distance between them is $4$. So if we take $\delta=4$ or any smaller positive number, then $0\lt|x-24|\lt4$, meaning all $x$ would be placed in the interval $(20,30)$ so that $|\frac{120}{x}-5|\lt1$. In other words, $$0\lt |x-24|\lt4\Rightarrow|\frac{120}{x}-5|\lt1$$

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