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: 34


1) $L=\lim_{x\to-5}f(x)=-4$ 2) $\delta=0.05$

Work Step by Step

$$f(x)=\frac{x^2+6x+5}{x+5}\hspace{1cm}c=-5\hspace{1cm}\epsilon=0.05$$ 1) Find $L=\lim_{x\to-5}f(x)$ $$\lim_{x\to-5}f(x)=\lim_{x\to-5}\frac{x^2+6x+5}{x+5}=\lim_{x\to-5}\frac{(x+1)(x+5)}{x+5}=\lim_{x\to-5}(x+1)$$ $$\lim_{x\to-5}f(x)=-5+1=-4$$ Therefore, $L=\lim_{x\to-5}f(x)=-4$ 2) Find $\delta\gt0$ such that for all $x$, $0\lt |x-(-5)|\lt\delta$ then $|f(x)-(-4)|\lt0.05$ - First, solve the inequality to figure out the open interval on which the inequality holds. $$|f(x)-(-4)|\lt0.05$$ $$|\frac{x^2+6x+5}{x+5}+4|\lt0.05$$ $$|\frac{x^2+6x+5+4x+20}{x+5}|\lt0.05$$ $$|\frac{x^2+10+25x}{x+5}|\lt0.05$$ $$|\frac{(x+5)^2}{x+5}|\lt0.05$$ $$|x+5|\lt0.05$$ $$-0.05\lt x+5\lt0.05$$ $$-5.05\lt x\lt-4.95$$ The open interval on which the inequality holds is $(-5.05,-4.95)$. - Find a value of $\delta\gt0$ The distance from $-5$ to both endpoints is the same, which is $0.05$. So if we take $\delta=0.05$, then $0\lt|x-(-5)|\lt0.05$, meaning all $x$ now are placed between $-5.05$ and $-4.95$, hence $|f(x)-(-4)|\lt0.05$ In other words, $$0\lt|x-(-5)|\lt0.05\Rightarrow|f(x)-(-4)|\lt0.05$$
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