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To solve the second limit, \( \lim_{x \to 2}(x^2 + 4x) = 12 \), using a similar structure to the first proof, we will proceed with an \(\epsilon\)-\(\delta\) proof. Here's how to do it step by step:

### Step 1: Define the limit

We want to show that:

\[
\lim_{x \to 2} (x^2 + 4x) = 12
\]

Formally, for every \(\epsilon > 0\), there must exist a \(\delta > 0\) such that if \(0 < |x - 2| < \delta\), then:

\[
|x^2 + 4x - 12| < \epsilon
\]

### Step 2: Rewrite the function in terms of \(x - 2\)

Start by simplifying the expression:

\[
|x^2 + 4x - 12| = |x^2 + 4x - 8 - 4| = |(x-2)(x+6)|
\]

We want to make this expression smaller than \(\epsilon\), i.e.,

\[
|(x-2)(x+6)| < \epsilon
\]

### Step 3: Bound the term \(x + 6\)

We need to control \(x + 6\) by keeping \(x\) close to 2. Specifically, if we assume \(|x - 2| < 1\), this implies:

\[
1 < x < 3
\]

Thus:

\[
7 < x + 6 < 9
\]

So we can bound \(x + 6\) by 9, i.e., \(x + 6 \leq 9\).

### Step 4: Establish the relationship between \(\delta\) and \(\epsilon\)

Now, we have:

\[
|(x - 2)(x + 6)| < \epsilon
\]

Given that \(x + 6 \leq 9\), it follows that:

\[
|x - 2| \cdot 9 < \epsilon
\]

Or:

\[
|x - 2| < \frac{\epsilon}{9}
\]

### Step 5: Choose \(\delta\)

We choose \(\delta = \min\left(1, \frac{\epsilon}{9}\right)\).

### Step 6: Conclusion

If \(0 < |x - 2| < \delta\), then:

\[
|x^2 + 4x - 12| = |(x - 2)(x + 6)| < 9 |x - 2| < 9 \cdot \frac{\epsilon}{9} = \epsilon
\]

Thus, the limit is proved:

\[
\lim_{x \to 2} (x^2 + 4x) = 12
\]

---

Would you like further clarification on any step?

Here are 5 related questions to extend your understanding:

1. How does the \(\epsilon\)-\(\delta\) definition apply to other types of limits?
2. Can you apply this method to prove a limit where \(x \to \infty\)?
3. How would you prove limits of functions that involve trigonometric expressions?
4. What happens when a limit approaches from only one side (left-hand or right-hand)?
5. How would the proof change if the expression involved a cubic polynomial?

**Tip:** When working with the \(\epsilon\)-\(\delta\) definition, always bound the terms involving \(x\) to simplify the inequalities.

Using the epsilon-delta proof from the first limit, solve the second limit lim(x -> 2)(x^2 + 4x) = 12.

Learn how to solve the limit lim(x -> 2)(x^2 + 4x) = 12 using the epsilon-delta definition of limits. This proof involves factoring quadratic expressions and bounding terms to satisfy the epsilon-delta condition.

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