Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.9 The Formal Definition of a Limit - Exercises - Page 93: 19


$$\lim_{x\to 4}\sqrt{x}=2 $$

Work Step by Step

Let $\epsilon\gt0$ be given. We bound $|\sqrt{x}-2|$ by multiplying $\dfrac{\sqrt{x}+2}{\sqrt{x}+2}$ \begin{align*} |\sqrt{x}-2|&=\left|\sqrt{x}-2\left(\frac{\sqrt{x}+2}{\sqrt{x}+2}\right)\right|\\ &=\left|\frac{x-4}{\sqrt{x}+2}\right|\\ &=|x-4|\left|\frac{1}{\sqrt{x}+2}\right| \end{align*} assume $\delta\lt1,$ so that $|x-4|\lt1,$ and hence $\sqrt{x}+2\gt\sqrt{3}+2\gt3 .$ This gives us \begin{align*} |\sqrt{x}-2|&=|x-4|\left|\frac{1}{\sqrt{x}+2}\right|\\ &\lt|x-4| \frac{1}{3} \end{align*} Let $\delta=\min (1,3 \epsilon) .$ If $|x-4|\lt\delta$ $$ |\sqrt{x}-2|=|x-4|\left|\frac{1}{\sqrt{x}+2}\right|\lt|x-4| \frac{1}{3}\lt\delta \frac{1}{3}\lt3 \epsilon \frac{1}{3}=\epsilon $$ Hence $$\lim_{x\to 4}\sqrt{x}=2 $$
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