Calculus 8th Edition

Published by Cengage

Chapter 1 - Functions and Limits - 1.7 The Precise Definition of a Limit - 1.7 Exercises: 3

Answer

$\delta=1.44$ (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 1.6 and 2.4 $\sqrt{x}=1.6 \ \ \Rightarrow\ \ x=1.6^{2}=2.56$ $\sqrt{x}=2.4 \ \ \Rightarrow\ \ x=2.4^{2}=5.76$ So, the interval (to which x=4 belongs) is $(2.56, 5.76)$ or, rewritten as $I=(4-1.44, 4+1.76).$ From the graph, $x\in I \ \ \Rightarrow\ \ |f(x)-2| < 0.4$. (all values of $x\in I$ have f(x) within $\pm 0.4$ of 2) We take a subset from $I$, one that we can write as$(4-\delta,4+\delta)$, such as $I_{1}=(4-1.44,4+1.44)$ For $x\in I_{1}\ \ \Rightarrow\ \ x\in I\ \ \Rightarrow\ \ |f(x)-2| < 0.4$. So for $\epsilon=0.4,$we have found $\delta=1.44$, such that $0 < |x-4| < \delta$, then $|f(x)-2| < \epsilon$.

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