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

Published by Pearson

# Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 76: 25

#### Answer

1) The open interval is $(\sqrt{15},\sqrt{17})$ 2) $\delta=\sqrt{17}-4$

#### Work Step by Step

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-4|\lt\delta\Rightarrow|(x^2-5)-11|\lt1$$ 1) Find the interval around $4$ on which $|(x^2-5)-11|\lt1$ holds. Solve the inequality: $$|(x^2-5)-11|\lt1$$ $$-1\lt(x^2-5)-11\lt1$$ $$10\lt x^2-5\lt12$$ $$15\lt x^2\lt17$$ Here we examine the interval around $4$, so we automatically take $x$ to be greater than $0$. Therefore, $$\sqrt{15}\lt x\lt\sqrt{17}$$ The open interval around $4$ is $(\sqrt{15},\sqrt{17})$. 2) Give a value for $\delta$ The nearer endpoint to $4$ is $\sqrt{17}$, and the distance between them is $\sqrt{17}-4$. So if we take $\delta=\sqrt{17}-4$ or any smaller positive number, then $0\lt|x-4|\lt\sqrt{17}-4$, meaning all $x$ would be placed in the interval $(\sqrt{15},\sqrt{17})$ so that $|(x^2-5)-11|\lt1$. In other words, $$0\lt |x-4|\lt\sqrt{17}-4\Rightarrow|(x^2-5)-11|\lt1$$

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