Short Answer

For each limit statement , use algebra to find δ > 0 in terms of $ε$ > 0 so that if 0 < |x − c| < δ, then | f(x) − L| < $ε$ .
$\lim_{x \to 0} (x^{3} + 1) = 1$

$δ = ε^{\frac{1}{3}}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step1. Given information.

We have been given a limit statement as $\lim_{x \to 0} (x^{3} + 1) = 1$ .

We have to find $δ in terms of ε$ .

Step 2. Use algebra.

From the given limit statement, we can identify

$f (x) = x^{3} + 1 c = 0 L = 1 For ε > 0 |(x^{3} + 1) - 1| < ε |x^{3} + 1 - 1| < ε |x^{3}| < ε | x |^{3} < ε | x | < ε^{\frac{1}{3}} For 0 < | x - c | < δ, we get | x | < ε^{\frac{1}{3}} . Therefore, δ = ε^{\frac{1}{3}}$

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Calculus

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Short Answer

For each limit statement , use algebra to find δ > 0 in terms of $ε$ > 0 so that if 0 < |x − c| < δ, then | f(x) − L| < $ε$ .
$\lim_{x \to 0} (x^{3} + 1) = 1$

Step by Step Solution

Step1. Given information.

Step 2. Use algebra.

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Calculus

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For each limit statement , use algebra to find δ > 0 in terms of ε > 0 so that if 0 < |x − c| < δ, then | f(x) − L| < ε.limx→0(x3+1)=1

Step by Step Solution

Step1. Given information.

Step 2. Use algebra.

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For each limit statement , use algebra to find δ > 0 in terms of $ε$ > 0 so that if 0 < |x − c| < δ, then | f(x) − L| < $ε$ .
$\lim_{x \to 0} (x^{3} + 1) = 1$