Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.7 The Precise Definition of a Limit - 1.7 Exercises: 4

Answer

Sample answer $\delta=0.1$ (or any smaller positive number)

Work Step by Step

First, we address the question marks on the x-axis. They stand for the numbers whose function values are $0.5$ and $1.5$ $x^{2}=0.5 \ \ \Rightarrow\ \ x=\sqrt{0.5}\approx$0.707106781187 $\sqrt{x}=2.4 \ \ \Rightarrow\ \ x=\sqrt{1.5}\approx$1.22474487139 So, the interval (to which x=$1$ belongs) is $I=(\sqrt{0.5}, \sqrt{1.5})$ . From the graph, $x\in I \displaystyle \ \ \Rightarrow\ \ |f(x)-1| < \frac{1}{5}$. (all values of $x\in I$ have f(x) within $\displaystyle \pm\frac{1}{5} $ of $1$) We take a subset from $I$, one that we can write as$ (1-\delta,1+\delta)$, such as $I_{1}=(1-0.1, 1+0.1)$ (the problem asks to find a $\delta$, not the greatest possible $\delta$, which would be $\sqrt{1.5}-1$, as $\sqrt{1.5}$ is closer to 1 than $\sqrt{0.5}$) For $x\displaystyle \in I_{1}\ \ \Rightarrow\ \ x\in I\ \ \Rightarrow\ \ |f(x)-1| < \frac{1}{2}$. So for $\displaystyle \epsilon=\frac{1}{2}, $we have found a number, $\delta=0.1$, such that $ 0 < |x-1| < \delta$, then $|x^{2}-1| < \epsilon$.
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