Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.2 Exercises - Page 56: 33


$$L=8$$ and $$\delta=\frac{1}{300}$$

Work Step by Step

1) From the definition of limit, the statement $$\lim_{x\to c}f(x)=L$$ means for each $\varepsilon\gt0$ there exists a $\delta\gt0$ such that if $$0\lt|x-c|\lt\delta$$ then $$|f(x)-L|\lt\varepsilon$$ 2) $$\lim_{x\to2}(3x+2)$$ To find $L$, we plug $x=2$ into $f(x)$ $$\lim_{x\to2}(3x+2)=3\times2+2=8$$ So $L=8$. Next, the question asks to find $\delta\gt0$ such that if $|f(x)-L|\lt0.01$ whenever $0\lt|x-c|\lt\delta$. Here, $c=2$, $L=8$ and $f(x)=3x+2$. Therefore, $$|x-c|=|x-2|$$ and $$|f(x)-L|=|3x+2-8|=|3x-6|=3|x-2|$$ Since $|f(x)-L|\lt0.01$, that means $3|x-2|\lt0.01$ Therefore as we need to find $\delta$ for $0\lt|x-2|\lt\delta$, we can pick $\delta=\frac{0.01}{3}=\frac{1}{300}$. 3) Try $\delta=\frac{1}{300}$ back again For $0\lt|x-2|\lt\frac{1}{300}$, we have $$|f(x)-L|=3|x-2|\lt\Big(3\times\frac{1}{300}\Big)=\frac{1}{100}=0.01$$ So $\delta=\frac{1}{300}$ is a right choice.
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