## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 2: Limits and Continuity - Section 2.3 - The Precise Definition of a Limit - Exercises 2.3 - Page 66: 18

#### Answer

$\left(\dfrac{4}{25}, \dfrac{9}{25}\right)$ $\delta=\dfrac{9}{100}$

#### Work Step by Step

We are given $f(x)=\sqrt{x}$, $L=\dfrac{1}{2}$, $c=\dfrac{1}{4}$, and $\epsilon=0.1=\dfrac{1}{10}$. To find the desired interval about $c$, we plug the above into $|f(x)-L| \lt \epsilon$ and solve for $x$: $$\left|\sqrt{x}-\dfrac{1}{2} \right| \lt \dfrac{1}{10}\\ -\dfrac{1}{10}\lt \sqrt{x}-\dfrac{1}{2} \lt \dfrac{1}{10}\\ \dfrac{4}{10}\lt \sqrt{x} \lt \dfrac{6}{10}\\ \dfrac{2}{5}\lt \sqrt{x} \lt \dfrac{3}{5}\\ \dfrac{4}{25}\lt x \lt \dfrac{9}{25}.$$ Thus our interval is $\left(\dfrac{4}{25}, \dfrac{9}{25}\right)$. Now, to find our $\delta$, we note that $$\dfrac{4}{25} \lt x \lt \dfrac{9}{25} \implies \dfrac{4}{25}-\dfrac{1}{4} \lt x-\dfrac{1}{4} \lt \dfrac{9}{25}-\dfrac{1}{4}\\ \implies -\dfrac{9}{100} \lt x-\dfrac{1}{4} \lt \dfrac{11}{100}.$$ Hence for $\delta=\dfrac{9}{100}$, $$0 \lt \left|x-\dfrac{1}{4}\right| \lt \delta \implies \left|f(x)-\dfrac{1}{2}\right| \lt \epsilon.$$

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