Question
2. Proof (showing that πΏ works). Given π > 0, choose πΏ = π/2. If
0 <
Correct: Your answer is correct.
< πΏ,
then
|(2x β 5) β 7| =
Correct: Your answer is correct.
= 2
Correct: Your answer is correct.
< 2πΏ
= 2
Incorrect: Your answer is incorrect.
= π.
Thus
if 0 < |x β 6| < πΏ then |(2x β 5) β 7| < π.
Therefore, by the definition of a limit
lim
x β 6
2x β 5 = 7.
0 <
Correct: Your answer is correct.
< πΏ,
then
|(2x β 5) β 7| =
Correct: Your answer is correct.
= 2
Correct: Your answer is correct.
< 2πΏ
= 2
Incorrect: Your answer is incorrect.
= π.
Thus
if 0 < |x β 6| < πΏ then |(2x β 5) β 7| < π.
Therefore, by the definition of a limit
lim
x β 6
2x β 5 = 7.
Answers
GPT-4o mini
The steps you've laid out follow the formal epsilon-delta definition of a limit. However, there seems to be a small mistake in the part where you determined \( |(2x - 5) - 7| \).
Let's clarify the proof step by step:
1. **Goal**: We want to show that
\[
\lim_{x \to 6} (2x - 5) = 7.
\]
2. **Epsilon-Delta Definition**: According to the epsilon-delta definition, for every \( \epsilon > 0 \), we must find a \( \delta > 0 \) such that if \( 0 < |x - 6| < \delta \), then it follows that \( |(2x - 5) - 7| < \epsilon \).
3. **Set up the Expression**:
- First, simplify \( |(2x - 5) - 7| \):
\[
|(2x - 5) - 7| = |2x - 12| = 2|x - 6|.
\]
4. **Choose \( \delta \)**: Set \( \delta = \frac{\epsilon}{2} \).
5. **Show the Implication**: If \( 0 < |x - 6| < \delta \), then:
\[
|2x - 12| = 2|x - 6| < 2\delta.
\]
Substituting \( \delta = \frac{\epsilon}{2} \),
\[
2\delta = 2 \cdot \frac{\epsilon}{2} = \epsilon.
\]
Thus,
\[
|2x - 12| < \epsilon.
\]
6. **Conclude**: Therefore, for every \( \epsilon > 0 \), choosing \( \delta = \frac{\epsilon}{2} \) ensures that if \( 0 < |x - 6| < \delta \), then \( |(2x - 5) - 7| < \epsilon \). Hence, we conclude that:
\[
\lim_{x \to 6} (2x - 5) = 7.
\]
Your original proof was on the right track, but itβs important to ensure the transition steps are explicit to prevent confusion. Just remember to keep track of how \( |2x - 12| \) relates directly to \( |x - 6| \) and how choosing \( \delta \) influences this relationship.
Let's clarify the proof step by step:
1. **Goal**: We want to show that
\[
\lim_{x \to 6} (2x - 5) = 7.
\]
2. **Epsilon-Delta Definition**: According to the epsilon-delta definition, for every \( \epsilon > 0 \), we must find a \( \delta > 0 \) such that if \( 0 < |x - 6| < \delta \), then it follows that \( |(2x - 5) - 7| < \epsilon \).
3. **Set up the Expression**:
- First, simplify \( |(2x - 5) - 7| \):
\[
|(2x - 5) - 7| = |2x - 12| = 2|x - 6|.
\]
4. **Choose \( \delta \)**: Set \( \delta = \frac{\epsilon}{2} \).
5. **Show the Implication**: If \( 0 < |x - 6| < \delta \), then:
\[
|2x - 12| = 2|x - 6| < 2\delta.
\]
Substituting \( \delta = \frac{\epsilon}{2} \),
\[
2\delta = 2 \cdot \frac{\epsilon}{2} = \epsilon.
\]
Thus,
\[
|2x - 12| < \epsilon.
\]
6. **Conclude**: Therefore, for every \( \epsilon > 0 \), choosing \( \delta = \frac{\epsilon}{2} \) ensures that if \( 0 < |x - 6| < \delta \), then \( |(2x - 5) - 7| < \epsilon \). Hence, we conclude that:
\[
\lim_{x \to 6} (2x - 5) = 7.
\]
Your original proof was on the right track, but itβs important to ensure the transition steps are explicit to prevent confusion. Just remember to keep track of how \( |2x - 12| \) relates directly to \( |x - 6| \) and how choosing \( \delta \) influences this relationship.