Chapter 1: Q. 32 (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| < $ε$ .
$\lim_{x \to 1} (3 x + 8) = 11$

Short Answer

Expert verified

$δ = \frac{ε}{3}$

Step by step solution

Step1. Given information.

We have been given a limit statement as $\lim_{x \to 1} (3 x + 8) = 11$ .

We have to find $δ in terms of ε$ .

Step 2. Use algebra

From the given limit statement, we can identify

$f (x) = 3 x + 8 c = 1 L = 11 For ε > 0 | (3 x + 8) - 11 | < ε | 3 x + 8 - 11 | < ε | 3 x - 3 | < ε 3 | x - 1 | < ε | x - 1 | < \frac{ε}{3} For 0 < | x - c | < δ, we get | x - 1 | < \frac{ε}{3} Therefore, δ = \frac{ε}{3}$

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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| < $ε$ .
$\lim_{x \to 1} (3 x + 8) = 11$

Short Answer

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.limx→1(3x+8)=11

Short Answer

Step by step solution

Step1. Given information.

Step 2. Use algebra

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Discrete Mathematics

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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| < $ε$ .
$\lim_{x \to 1} (3 x + 8) = 11$