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

Published by Pearson

# Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 76: 13

#### Answer

$\delta=9/25$

#### Work Step by Step

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-(-1)|\lt\delta\Rightarrow|f(x)-2|\lt0.5$$ Looking at the graphs, we can see that for values of $f(x)$ to be restricted between $1.5$ and $2.5$ (which means $|f(x)-2|\lt0.5$), $x$ must be placed between $-\frac{16}{9}$ and $-\frac{16}{25}$. The graph also shows that the endpoint $-\frac{16}{25}$ is nearer to $-1$ than $-\frac{16}{25}$, so we would take its distance from $-1$ to $-\frac{16}{25}$ to be $\delta$. $$\delta=|-1-(-\frac{16}{25})|=|-1+\frac{16}{25}|=|\frac{16-25}{25}|=|-\frac{9}{25}|=\frac{9}{25}$$ Therefore, $$0\lt |x-(-1)|\lt\frac{9}{25}\Rightarrow|f(x)-2|\lt0.5$$

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