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: 35


1) $L=\lim_{x\to-3}f(x)=4$ 2) $\delta=0.75$

Work Step by Step

$$f(x)=\sqrt{1-5x}\hspace{1cm}c=-3\hspace{1cm}\epsilon=0.5$$ 1) Find $L=\lim_{x\to-3}f(x)$ $$\lim_{x\to-3}f(x)=\lim_{x\to-3}\sqrt{1-5x}=\sqrt{1-5\times(-3)}=\sqrt{1+15}=\sqrt{16}=4$$ Therefore, $L=\lim_{x\to-3}f(x)=4$ 2) Find $\delta\gt0$ such that for all $x$, $0\lt |x-(-3)|\lt\delta$ then $|f(x)-4|\lt0.5$ - First, solve the inequality to figure out the open interval on which the inequality holds. $$|f(x)-4|\lt0.5$$ $$|\sqrt{1-5x}-4|\lt0.5$$ $$-0.5\lt\sqrt{1-5x}-4\lt0.5$$ $$3.5\lt\sqrt{1-5x}\lt4.5$$ Square: $$12.25\lt1-5x\lt20.25$$ $$11.25\lt-5x\lt19.25$$ $$-2.25\gt x\gt-3.85$$ The open interval on which the inequality holds is $(-3.85,-2.25)$. - Find a value of $\delta\gt0$ $-3$ is nearer to the endpoint $-2.25$, whose distance is $0.75$. So if we take $\delta=0.75$, then $0\lt|x-(-3)|\lt0.75$, meaning all $x$ now are placed between $-3.85$ and $-2.25$, hence $|f(x)-4|\lt0.5$ In other words, $$0\lt|x-(-3)|\lt0.75\Rightarrow|f(x)-4|\lt0.5$$
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