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Expert-verified Found in: Page 107 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # For each limit statement , use algebra to find δ > 0 in terms of $\epsilon$ > 0 so that if 0 < |x − c| < δ, then | f(x) − L| < $\epsilon$.$\underset{x\to 0}{\mathrm{lim}}\left({x}^{3}+1\right)=1$

$\delta ={\epsilon }^{\frac{1}{3}}$

See the step by step solution

## Step1. Given information.

We have been given a limit statement as $\underset{x\to 0}{\mathrm{lim}}\left({x}^{3}+1\right)=1$.

We have to find $\delta \mathrm{in}\mathrm{terms}\mathrm{of}\epsilon$.

## Step 2. Use algebra.

From the given limit statement, we can identify

$f\left(x\right)={x}^{3}+1\phantom{\rule{0ex}{0ex}}c=0\phantom{\rule{0ex}{0ex}}L=1\phantom{\rule{0ex}{0ex}}\mathrm{For}\epsilon >0\phantom{\rule{0ex}{0ex}}\left|\left({x}^{3}+1\right)-1\right|<\epsilon \phantom{\rule{0ex}{0ex}}\left|{x}^{3}+1-1\right|<\epsilon \phantom{\rule{0ex}{0ex}}\left|{x}^{3}\right|<\epsilon \phantom{\rule{0ex}{0ex}}|x{|}^{3}<\epsilon \phantom{\rule{0ex}{0ex}}|x|<{\epsilon }^{\frac{1}{3}}\phantom{\rule{0ex}{0ex}}\mathrm{For}0<|x-c|<\delta ,\mathrm{we}\mathrm{get}|x|<{\epsilon }^{\frac{1}{3}}.\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\delta ={\epsilon }^{\frac{1}{3}}$ ### Want to see more solutions like these? 