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Answers without the blur. Sign up and see all textbooks for free! Q. 54

Expert-verified Found in: Page 98 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # For each limit $\underset{x\to c}{lim}f\left(x\right)=L$in Exercises 43–54, use graphs and algebra to approximate the largest value of $\delta$ such that if $x\in \left(c-\delta ,c\right)\cup \left(c,c+\delta \right)\mathrm{then}\mathrm{f}\left(\mathrm{x}\right)\in \left(\mathrm{L}-\epsilon ,\mathrm{L}+\epsilon \right)\mathit{}$$\underset{x\to -1}{lim}\frac{{x}^{2}-2x-3}{x+1}=-4,\epsilon =0.1$

The required value of $\delta =0.1$

See the step by step solution

## Step 1. Given Information

The given expression is $\underset{x\to -1}{lim}\frac{{x}^{2}-2x-3}{x+1}=-4,\epsilon =0.1$

## Step 2. Explanation

From the given expression, we have, $c=-1,L=-4$

The limit expression can be written as a formal statement as below,

For all epsilon positive, there exists a delta positive such that if $x\in \left(-1-\delta ,-1\right)\cup \left(-1,-1+\delta \right)\mathrm{Then}\mathit{}\frac{{\mathrm{x}}^{2}-2\mathrm{x}-3}{\mathrm{x}+1}\in \left(-4-\mathrm{\epsilon },-4+\mathrm{\epsilon }\right)$

Now, the largest value of delta is given by,

$\frac{{x}^{2}-2x-3}{x+1}=L+\epsilon \phantom{\rule{0ex}{0ex}}\frac{{x}^{2}-2x-3}{x+1}=-4+1\phantom{\rule{0ex}{0ex}}x-3=-3\phantom{\rule{0ex}{0ex}}x=0$

Thus,

$\delta =0+0.1\phantom{\rule{0ex}{0ex}}\delta =0.1$ ### Want to see more solutions like these? 