Chapter 1: Q. 60 (page 108)

Write delta-epsilon proofs for each of the limit statements in Exercises $47 - 60 .$
$\lim_{x \to 2^{+}} 3 \sqrt{2 x - 4} = 0$

Short Answer

Expert verified

Delta-epsilon proof is,

Whenever $x \in (2, 2 + δ)$ , we also have $| 3 \sqrt{2 x - 4} - 0 | < ϵ$ .

Step by step solution

Step 1. Given information

We are given,

$\lim_{x \to 2^{+}} 3 \sqrt{2 x - 4} = 0$

Step 2. Writing the delta-epsilon proofs

The strategy is to write delta-epsilon proofs for the given limit statement.

Consider that $\in > 0$ , choose $δ = \frac{ϵ^{2}}{18}$ .

The limit statement $\lim_{x \to 2^{+}} 3 \sqrt{2 x - 4} = 0$ means that for all $\in > 0$ , there exist $δ > 0$ such that if $x \in (2, 2 + δ)$ , then $| 3 \sqrt{2 x - 4} - 0 | < ϵ$ .

So,

role="math" localid="1648029573923" $2 < x < 2 + δ 2 - 2 < x - 2 < 2 + δ - 2 0 < x - 2 < δ$

This means that, when $0 < x - 2 < δ$ , we have

$| 3 \sqrt{2 x - 4} - 0 | = | 3 \sqrt{2 x - 4} | = 3 \sqrt{2 (x - 2)} < \sqrt{18} \sqrt{δ} = \sqrt{18} \sqrt{\frac{\in^{2}}{18}} = \in$

So, whenever $x \in (2, 2 + δ)$ , we also have $| 3 \sqrt{2 x - 4} - 0 | < ϵ$ .

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Write delta-epsilon proofs for each of the limit statements in Exercises $47 - 60 .$
$\lim_{x \to 2^{+}} 3 \sqrt{2 x - 4} = 0$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Writing the delta-epsilon proofs

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Write delta-epsilon proofs for each of the limit statements in Exercises 47–60.limx→2+32x-4=0

Short Answer

Step by step solution

Step 1. Given information

Step 2. Writing the delta-epsilon proofs

One App. One Place for Learning.

Most popular questions from this chapter

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Write delta-epsilon proofs for each of the limit statements in Exercises $47 - 60 .$
$\lim_{x \to 2^{+}} 3 \sqrt{2 x - 4} = 0$