Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Use Epsilon Delta Definition to prove lim x goes to 2, x^3 = 8
![**Using the ε-δ definition to prove the limit:**
**Problem Statement:**
Use the ε-δ definition to prove \(\lim_{{x \to 2}} x^3 = 8\).
**Explanation:**
The ε-δ definition of a limit states that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < |x - c| < \delta\), then \(|f(x) - L| < \epsilon\). Here, \(c = 2\) and the function \(f(x) = x^3\) with the limit \(L = 8\).
**Proof:**
To show:
\[
\forall \epsilon > 0, \exists \delta > 0 \text{ such that } 0 < |x - 2| < \delta \Rightarrow |x^3 - 8| < \epsilon
\]
1. **Start with the goal:**
\[
|x^3 - 8| < \epsilon
\]
Since:
\[
|x^3 - 8| = |(x - 2)(x^2 + 2x + 4)|
\]
We need this to be:
\[
|x - 2| |x^2 + 2x + 4| < \epsilon
\]
2. **Finding Bounds for \( |x^2 + 2x + 4| \):**
Choose a small enough \(|x - 2| < 1\), hence, \(1 < x < 3\).
3. **Bounding \( |x^2 + 2x + 4| \) when \(1 < x < 3\):**
When \(1 < x < 3\):
\[
x^2 < 9, \quad 2x < 6, \quad \text{and} \quad 4 \text{ is constant}
\]
So,
\[
|x^2 + 2x + 4| < 9 + 6 + 4 = 19
\]
Thus:
\[
|x^2 + 2x + 4| \leq 19
\]
4. **Relating \(|x-2|\) and \(\epsilon\):**
Given:
\[
|x -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc88a8d3c-414f-44f0-8ef5-044fbd549dcd%2Fe58b0274-ea31-46aa-8fe6-ff96e1286679%2F6p2bio.png&w=3840&q=75)
Transcribed Image Text:**Using the ε-δ definition to prove the limit:**
**Problem Statement:**
Use the ε-δ definition to prove \(\lim_{{x \to 2}} x^3 = 8\).
**Explanation:**
The ε-δ definition of a limit states that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < |x - c| < \delta\), then \(|f(x) - L| < \epsilon\). Here, \(c = 2\) and the function \(f(x) = x^3\) with the limit \(L = 8\).
**Proof:**
To show:
\[
\forall \epsilon > 0, \exists \delta > 0 \text{ such that } 0 < |x - 2| < \delta \Rightarrow |x^3 - 8| < \epsilon
\]
1. **Start with the goal:**
\[
|x^3 - 8| < \epsilon
\]
Since:
\[
|x^3 - 8| = |(x - 2)(x^2 + 2x + 4)|
\]
We need this to be:
\[
|x - 2| |x^2 + 2x + 4| < \epsilon
\]
2. **Finding Bounds for \( |x^2 + 2x + 4| \):**
Choose a small enough \(|x - 2| < 1\), hence, \(1 < x < 3\).
3. **Bounding \( |x^2 + 2x + 4| \) when \(1 < x < 3\):**
When \(1 < x < 3\):
\[
x^2 < 9, \quad 2x < 6, \quad \text{and} \quad 4 \text{ is constant}
\]
So,
\[
|x^2 + 2x + 4| < 9 + 6 + 4 = 19
\]
Thus:
\[
|x^2 + 2x + 4| \leq 19
\]
4. **Relating \(|x-2|\) and \(\epsilon\):**
Given:
\[
|x -
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