To demonstrate that $\lim_{x \to 2} (2x + 2) = 6$ with $\epsilon = 0.2$, we need to find a value of $\delta$ that satisfies the definition of a limit:

$\text{For all } x, \text{ if } 0 < |x - 2| < \delta, \text{ then } |(2x + 2) - 6| < 0.2.$

Step-by-step:

1. Start with the inequality: $|(2x + 2) - 6| < 0.2.$

2. Simplify: $|2x + 2 - 6| = |2x - 4| = 2|x - 2|.$

   So we need: $2|x - 2| < 0.2.$

3. Solve for $|x - 2|$: $|x - 2| < \frac{0.2}{2} = 0.1.$

Thus, a value of $\delta = 0.1$ suffices to ensure that $|(2x + 2) - 6| < 0.2$.

Conclusion:

For $\epsilon = 0.2$, $\delta = 0.1$ is sufficient to demonstrate that $\lim_{x \to 2} (2x + 2) = 6$.

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Do you want further details or have any questions?

Here are 5 related questions:

1. How would you find $\delta$ for a different $\epsilon$ value, such as $\epsilon = 0.1$?
2. Can you explain how the formal definition of a limit works?
3. What happens if you use a larger $\delta$ value, like $\delta = 0.2$?
4. How can this limit be visualized on a graph of the function $f(x) = 2x + 2$?
5. What are the key steps in finding $\delta$ for any general limit problem?

Tip: Always simplify the inequality after substituting the limit into the function to find $\delta$.