Chapter 1: Q. 39 (page 107)

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| < $ε$ .
$\lim_{x \to 3} (x^{2} - 2 x - 3) = 0; you may assume δ \leq 1$

Short Answer

Expert verified

$δ = m i n (1, \frac{ε}{5})$

Step by step solution

Step1. Given information.

We have been given a limit statement as $\lim_{x \to 3} (x^{2} - 2 x - 3) = 0; you may assume δ \leq 1$ .

We have to find $δ in terms of ε$ .

Step 2. Use algebra.

From the given limit statement, we can identify

$f (x) = x^{2} - 2 x - 3 c = 3 L = 0 For ε > 0 |(x^{2} - 2 x - 3) - 0| < ε |x^{2} - 3 x + x - 3| < ε | x (x - 3) + 1 (x - 3) | < ε | (x - 3) (x + 1) | < ε | x - 3 | \cdot | x + 1 | < ε δ | x + 1 | < ε$

Step 3. Put x=4

$δ \cdot | 4 + 1 | < ε δ \cdot | 5 | < ε δ < \frac{ε}{5} For 0 < | x - c | < δ, we get | x - 3 | < \frac{ε}{5} Therefore, δ = m i n (1, \frac{ε}{5})$

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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| < $ε$ .
$\lim_{x \to 3} (x^{2} - 2 x - 3) = 0; you may assume δ \leq 1$

Short Answer

Step by step solution

Step1. Given information.

Step 2. Use algebra.

Step 3. Put x=4

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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→3(x2-2x-3)=0;youmayassumeδ≤1

Short Answer

Step by step solution

Step1. Given information.

Step 2. Use algebra.

Step 3. Put x=4

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Decision Maths

Calculus

Discrete Mathematics

Probability and Statistics

Statistics

Study anywhere. Anytime. Across all devices.

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| < $ε$ .
$\lim_{x \to 3} (x^{2} - 2 x - 3) = 0; you may assume δ \leq 1$