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Q. 43

Expert-verifiedFound in: Page 97

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 $x\in (c-\delta ,c)\cup (c,c+\delta )\mathrm{then}\mathrm{f}\left(\mathrm{x}\right)\in (\mathrm{L}-\epsilon ,\mathrm{L}+\epsilon )\mathit{}$.

role="math" localid="1648027918805" $\underset{\mathrm{x}\to 2}{\mathrm{lim}}{\mathrm{x}}^{3}=8,\mathrm{\epsilon}=0.5$

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

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

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

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,

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

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

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