University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 76: 36


1) $L=\lim_{x\to2}f(x)=2$ 2) $\delta=1/3$

Work Step by Step

$$f(x)=\frac{4}{x}\hspace{1cm}c=2\hspace{1cm}\epsilon=0.4$$ 1) Find $L=\lim_{x\to2}f(x)$ $$\lim_{x\to2}f(x)=\lim_{x\to2}\frac{4}{x}=\frac{4}{2}=2$$ Therefore, $L=\lim_{x\to2}f(x)=2$ 2) Find $\delta\gt0$ such that for all $x$, $0\lt |x-2|\lt\delta$ then $|f(x)-2|\lt0.4$ - First, solve the inequality to figure out the open interval on which the inequality holds. $$|f(x)-2|\lt0.4$$ $$|\frac{4}{x}-2|\lt0.4$$ $$-0.4\lt\frac{4}{x}-2\lt0.4$$ $$1.6\lt\frac{4}{x}\lt2.4$$ $$\frac{4}{1.6}\gt x\gt \frac{4}{2.4}$$ $$2.5\gt x\gt\frac{5}{3}$$ The open interval on which the inequality holds is $(5/3,2.5)$. - Find a value of $\delta\gt0$ $2$ is nearer to the endpoint $5/3$, whose distance is $1/3$. So if we take $\delta=1/3$, then $0\lt|x-2|\lt1/3$, meaning all $x$ now are placed between $5/3$ and $2.5$, hence $|f(x)-2|\lt0.4$ In other words, $$0\lt|x-2|\lt\frac{1}{3}\Rightarrow|f(x)-2|\lt0.4$$
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