## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 76: 14

#### Answer

$\delta=0.0024$

#### Work Step by Step

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-\frac{1}{2}|\lt\delta\Rightarrow|f(x)-2|\lt0.01$$ Looking at the graphs, we can see that for values of $f(x)$ to be restricted between $1.99$ and $2.01$ (which means $|f(x)-2|\lt0.01$), $x$ must be placed between $\frac{1}{2.01}$ and $\frac{1}{1.99}$. We need to calculate the distance from $\frac{1}{2}$ to $\frac{1}{2.01}$ and $\frac{1}{1.99}$: $\frac{1}{2}-\frac{1}{2.01}\approx0.00249$ and $\frac{1}{1.99}-\frac{1}{2}\approx0.00251$ Since the distances are not exact, to make sure $x$ stay within the endpoints, we can choose $\delta=0.0024$ Therefore, $$0\lt |x-\frac{1}{2}|\lt0.0024\Rightarrow|f(x)-2|\lt0.01$$

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