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 75: 9

Answer

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-1|\lt\delta\Rightarrow|f(x)-1|\lt\frac{1}{4}$$ Looking at the graphs, we can see that for values of $f(x)$ to be restricted between $\frac{3}{4}$ and $\frac{5}{4}$ (which means $|f(x)-1|\lt\frac{1}{4}$), $x$ must be placed between $\frac{9}{16}$ and $\frac{25}{16}$. $|\frac{25}{16}-1|=|\frac{9}{16}|=\frac{9}{16}$ And, $|1-\frac{9}{16}|=|\frac{7}{16}|=\frac{7}{16}$ Since, $\frac{7}{16} \lt \frac{9}{16},$ $\delta=\frac{7}{16}$

Work Step by Step

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-1|\lt\delta\Rightarrow|f(x)-1|\lt\frac{1}{4}$$ Looking at the graphs, we can see that for values of $f(x)$ to be restricted between $\frac{3}{4}$ and $\frac{5}{4}$ (which means $|f(x)-1|\lt\frac{1}{4}$), $x$ must be placed between $\frac{9}{16}$ and $\frac{25}{16}$. And since $|\frac{25}{16}-1|=|\frac{9}{16}|=\frac{9}{16}$, that means $0\lt |x-1|\lt\frac{9}{16}$. Therefore, $\delta=\frac{9}{16}$ (or any smaller positives number is also fine).
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