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Q. 43

Expert-verified
Found in: Page 97

### 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 $x\in \left(c-\delta ,c\right)\cup \left(c,c+\delta \right)\mathrm{then}\mathrm{f}\left(\mathrm{x}\right)\in \left(\mathrm{L}-\epsilon ,\mathrm{L}+\epsilon \right)\mathit{}$.role="math" localid="1648027918805" $\underset{\mathrm{x}\to 2}{\mathrm{lim}}{\mathrm{x}}^{3}=8,\mathrm{\epsilon }=0.5$

The largest value of $\delta \approx 0.0408$.

See the step by step solution

## Step 1. Given Information.

The given expression is $\underset{\mathrm{x}\to 2}{\mathrm{lim}}{\mathrm{x}}^{3}=8,\mathrm{\epsilon }=0.5$.

## Step 2. Explanation.

From the given expression, we have, $c=2,L=8$.

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

For all epsilon positive, there exists a delta positive such that if

localid="1648188825671" $x\in \left(2-\delta ,2\right)\cup \left(2,2+\delta \right)\mathrm{Then}\mathit{}{x}^{\mathit{3}}\in \left(8-\mathrm{\epsilon },8+\mathrm{\epsilon }\right)$

Now, the largest value of delta is given by,

${x}^{3}=L+\epsilon \phantom{\rule{0ex}{0ex}}{x}^{3}=8+0.5\phantom{\rule{0ex}{0ex}}{x}^{3}=8.5\phantom{\rule{0ex}{0ex}}x={\left(8.5\right)}^{\frac{1}{3}}$

Thus,

$\delta ={\left(8.5\right)}^{\frac{1}{3}}-2\phantom{\rule{0ex}{0ex}}\delta \approx 0.0408$

## Step 3. Conclusion.

The largest value of $\delta \approx 0.0408$.