4. Use the e, 8 definition to show that lim,- (4x + 5) = 9.
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Numerical Analysis
![**Problem 4:**
Use the \( \epsilon, \delta \) definition to show that \( \lim_{{x \to 1}} (4x + 5) = 9 \).
**Solution:**
To demonstrate this using the \( \epsilon, \delta \) definition of a limit, we need to show that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - 1| < \delta \), then \( |(4x + 5) - 9| < \epsilon \).
Let's go through the process:
1. Start with the expression \( |(4x + 5) - 9| \).
2. Simplify:
\[
|4x + 5 - 9| = |4x - 4| = 4|x - 1|
\]
3. We require \( 4|x - 1| < \epsilon \).
4. Solve for \( |x - 1| \):
\[
|x - 1| < \frac{\epsilon}{4}
\]
5. Set \( \delta = \frac{\epsilon}{4} \).
Now, for a given \( \epsilon > 0 \), if we choose \( \delta = \frac{\epsilon}{4} \), then whenever \( 0 < |x - 1| < \delta \), it follows that \( |(4x + 5) - 9| < \epsilon \).
This confirms that \( \lim_{{x \to 1}} (4x + 5) = 9 \) using the \( \epsilon, \delta \) definition of a limit.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19daf46f-68b0-4a9f-8350-e4bad1609c2a%2F602ec9f5-459b-431f-8906-c5c9e614b36b%2Ffbiwmlp_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 4:**
Use the \( \epsilon, \delta \) definition to show that \( \lim_{{x \to 1}} (4x + 5) = 9 \).
**Solution:**
To demonstrate this using the \( \epsilon, \delta \) definition of a limit, we need to show that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - 1| < \delta \), then \( |(4x + 5) - 9| < \epsilon \).
Let's go through the process:
1. Start with the expression \( |(4x + 5) - 9| \).
2. Simplify:
\[
|4x + 5 - 9| = |4x - 4| = 4|x - 1|
\]
3. We require \( 4|x - 1| < \epsilon \).
4. Solve for \( |x - 1| \):
\[
|x - 1| < \frac{\epsilon}{4}
\]
5. Set \( \delta = \frac{\epsilon}{4} \).
Now, for a given \( \epsilon > 0 \), if we choose \( \delta = \frac{\epsilon}{4} \), then whenever \( 0 < |x - 1| < \delta \), it follows that \( |(4x + 5) - 9| < \epsilon \).
This confirms that \( \lim_{{x \to 1}} (4x + 5) = 9 \) using the \( \epsilon, \delta \) definition of a limit.
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