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: 29

Answer

$\delta=0.4$

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) The question asks to find $\delta$ such that if $0\lt|x-2|\lt\delta$ then $|f(x)-3|\lt0.4$, with $f(x)=x+1$ So we have already been given $\varepsilon=0.4$. To find $\delta$, we need to find a connection between $|x-2|$ and $|f(x)-3|$. To do so, we replace $f(x)=x+1$ into $|f(x)-3|$. $$|f(x)-3|=|(x+1)-3|=|x-2|$$ Since $|f(x)-3|\lt0.4$, now $|x-2|\lt0.4$ Therefore as we need to find $\delta$ for $0\lt|x-2|\lt\delta$, we can pick $\delta=0.4$. 3) Try $\delta=0.4$ back again For $0\lt|x-2|\lt0.4$, we have $$|f(x)-3|=|(x+1)-3|=|x-2|\lt0.4$$ So $\delta=0.4$ is a right choice. *From the graph, we can also see that for x-values within $0.4$ of $2$ $(x\ne2)$, y-values are within $0.4$ of $3$.
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