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.