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The problem you've uploaded asks to find the value of \(\delta > 0\) such that if \( |x - 1| < \delta \), then \( |y - 5| < \epsilon \) where \( \epsilon = \frac{1}{2} \) and the function is \( y = 2x + 3 \).

### Step-by-step approach:

1. **Define the equation**: The function given is \( y = 2x + 3 \). At \( x = 1 \), we can find the value of \( y \):
   \[
   y(1) = 2(1) + 3 = 5
   \]
   So the point \( (1, 5) \) is the reference point for this limit problem.

2. **Restate the condition**: We are given the condition that \( |x - 1| < \delta \) and we need to ensure that \( |y - 5| < \epsilon \), where \( \epsilon = \frac{1}{2} \).

3. **Relate \( |y - 5| \) to \( |x - 1| \)**:
   \[
   y = 2x + 3 \quad \text{so} \quad |y - 5| = |2x + 3 - 5| = |2x - 2| = 2|x - 1|
   \]
   Thus, \( |y - 5| = 2|x - 1| \).

4. **Use the \( \epsilon \)-condition**: We need \( |y - 5| < \epsilon \). Using the relation \( |y - 5| = 2|x - 1| \), this becomes:
   \[
   2|x - 1| < \frac{1}{2}
   \]
   Divide both sides by 2:
   \[
   |x - 1| < \frac{1}{4}
   \]

5. **Conclusion**: This implies that \( \delta = \frac{1}{4} \), since \( \delta \) is the maximum value for which \( |x - 1| < \delta \) ensures \( |y - 5| < \epsilon \).

### Answer:
The correct answer is **(A) \(\delta = \frac{1}{4}\)** or smaller.

Would you like more details on this or have any questions?

Here are 5 related questions for further exploration:

1. How does the slope of a linear function affect the relationship between \(\delta\) and \(\epsilon\)?
2. Can this problem be generalized for any linear function \( y = mx + b \)?
3. What happens to \(\delta\) if \(\epsilon\) is made smaller or larger?
4. How would the problem change if the function were nonlinear, say \( y = x^2 \)?
5. What role does continuity play in this type of limit problem?

**Tip**: When solving \(\delta\)-\(\epsilon\) problems, always relate the change in \(y\) to the change in \(x\) by manipulating the functional relationship.

Given ε = 1/2, find all values of δ > 0 so that if |x - 1| < δ then |y - 5| < ε for the function y = 2x + 3.

Solve a delta-epsilon problem involving the linear function y = 2x + 3, where ε = 1/2. Learn how to find the appropriate δ that satisfies the given conditions for limits. Ideal for students studying calculus.

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