## 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: 17

#### Answer

$(-0.19,0.21)$ $\delta=0.19$

#### Work Step by Step

We are given $f(x)=\sqrt{x+1}$, $L=1$, $c=0$, and $\epsilon=0.1$. To find the desired interval about $c$, we plug the above into $|f(x)-L| \lt \epsilon$ and solve for $x$: $$\left|\sqrt{x+1}-1 \right| \lt 0.1\\ -0.1\lt \sqrt{x+1}-1 \lt 0.1 \\ 0.9\lt \sqrt{x+1} \lt1.1 \\ 0.81\lt x+1 \lt 1.21\\ -0.19 \lt x \lt 0.21.$$ Thus our interval is $(-0.19,0.21)$. Now, to find our $\delta$, we note that $$-0.19 \lt x-0 \lt 0.19 \implies -0.19 \lt x-0 \lt 0.21.$$ Hence for $\delta=0.19$, $$0 \lt |x-0| \lt \delta \implies |f(x)-1| \lt \epsilon.$$

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