To prove that lim(x-7) 2x - 3= 11 using the ε- definition, we need there exists a 8 >0 such that if 0 < x - 7| <8, then |(2x-3) -11 | < | Given & > 0, we want to find a 8 >0 such that if 0 < x-7| <8, then [(2x-3) -11|
To prove that lim(x-7) 2x - 3= 11 using the ε- definition, we need there exists a 8 >0 such that if 0 < x - 7| <8, then |(2x-3) -11 | < | Given & > 0, we want to find a 8 >0 such that if 0 < x-7| <8, then [(2x-3) -11|
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Did I do this proof correctly?
![### Proving the Limit Using the ε-δ Definition
To prove that \(\lim_{{x \to 7}} 2x - 3 = 11\) using the ε-δ definition, we need to show that for any given \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - 7| < \delta \), then \( |2x - 3 - 11| < \varepsilon \).
#### Step-by-Step Solution:
1. **Restate the Problem:**
Given \( \varepsilon > 0 \), we want to find a \( \delta > 0 \) such that if \( 0 < |x - 7| < \delta \), then \( |2x - 3 - 11| < \varepsilon \).
2. **Simplify the Expression:**
Let's work with the expression \(|2x - 3 - 11|\) and simplify it:
\[
|2x - 3 - 11| = |2x - 14|
\]
3. **Finding an Upper Bound:**
Now, we can try to find an upper bound for \( |2x - 14| \) based on a given \( \varepsilon \).
4. **Rewriting the Expression:**
We can rewrite \( |2x - 14| \) as \( |2(x - 7)| \). Notice that when \( |x - 7| < 1 \), we have:
\[
|2(x - 7)| < 2
\]
5. **Choosing \( \delta \):**
So, if we choose \( \delta = 1 \), then for any \( 0 < |x - 7| < 1 \), we have \( |2x - 14| < 2 \).
6. **Proceed to Show \( |2x - 14| < \varepsilon \):**
Now, we can proceed to show that \( |2x - 3 - 11| < \varepsilon \) when \( 0 < |x - 7| < \delta \):
\[
|2x - 14| = |2(x - 7)| < 2
\]
We can choose](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd67763fa-2598-4b36-80a7-0b76d34ef205%2F1dc58bda-6c2c-4b45-884f-ed2776f63935%2Fo77lv5j_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Proving the Limit Using the ε-δ Definition
To prove that \(\lim_{{x \to 7}} 2x - 3 = 11\) using the ε-δ definition, we need to show that for any given \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - 7| < \delta \), then \( |2x - 3 - 11| < \varepsilon \).
#### Step-by-Step Solution:
1. **Restate the Problem:**
Given \( \varepsilon > 0 \), we want to find a \( \delta > 0 \) such that if \( 0 < |x - 7| < \delta \), then \( |2x - 3 - 11| < \varepsilon \).
2. **Simplify the Expression:**
Let's work with the expression \(|2x - 3 - 11|\) and simplify it:
\[
|2x - 3 - 11| = |2x - 14|
\]
3. **Finding an Upper Bound:**
Now, we can try to find an upper bound for \( |2x - 14| \) based on a given \( \varepsilon \).
4. **Rewriting the Expression:**
We can rewrite \( |2x - 14| \) as \( |2(x - 7)| \). Notice that when \( |x - 7| < 1 \), we have:
\[
|2(x - 7)| < 2
\]
5. **Choosing \( \delta \):**
So, if we choose \( \delta = 1 \), then for any \( 0 < |x - 7| < 1 \), we have \( |2x - 14| < 2 \).
6. **Proceed to Show \( |2x - 14| < \varepsilon \):**
Now, we can proceed to show that \( |2x - 3 - 11| < \varepsilon \) when \( 0 < |x - 7| < \delta \):
\[
|2x - 14| = |2(x - 7)| < 2
\]
We can choose
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