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The problem in the image asks to formally prove that:

\[
\lim_{x \to 6} \left( x^2 + x \right) = 42
\]

Given that \( \epsilon > 0 \) is arbitrary, and with a choice of \( \delta = \min\left( \frac{\epsilon}{m}, 1 \right) \), the task is to determine the smallest value of \( m \) that would satisfy the proof.

### Step-by-Step Approach:

1. **Limit Setup**: 
   To prove \( \lim_{x \to 6} (x^2 + x) = 42 \), we need to show that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( |x - 6| < \delta \), then \( |(x^2 + x) - 42| < \epsilon \).

2. **Simplify the Expression**:
   We start by simplifying \( |(x^2 + x) - 42| \):
   \[
   |(x^2 + x) - 42| = |x^2 + x - 42|
   \]
   Factor the expression:
   \[
   x^2 + x - 42 = (x - 6)(x + 7)
   \]
   Therefore:
   \[
   |x^2 + x - 42| = |(x - 6)(x + 7)|
   \]

3. **Establish a Bound for \( |x + 7| \)**:
   Since \( x \to 6 \), we know \( |x - 6| < 1 \). Thus, \( x \) lies between 5 and 7:
   \[
   5 < x < 7
   \]
   Therefore, \( x + 7 \) lies between 12 and 14, which gives us the bound:
   \[
   12 \leq |x + 7| \leq 14
   \]
   
4. **Choosing \( \delta \)**:
   Now we need to ensure that \( |(x - 6)(x + 7)| < \epsilon \). Using the bound on \( |x + 7| \), we have:
   \[
   |(x - 6)(x + 7)| \leq 14 |x - 6|
   \]
   To satisfy \( |(x - 6)(x + 7)| < \epsilon \), it is enough to have:
   \[
   14 |x - 6| < \epsilon \quad \text{or} \quad |x - 6| < \frac{\epsilon}{14}
   \]

5. **Conclusion**:
   Thus, we choose \( m = 14 \). The smallest value of \( m \) that satisfies the proof is 14.

### Final Result:
The smallest value of \( m \) that would satisfy the proof is \( m = 14 \).

---

Would you like more details or have any questions?

Here are five related questions to deepen your understanding:
1. How would the limit change if the function were \( x^2 + 2x \) instead of \( x^2 + x \)?
2. Can you explain why we need to bound \( |x + 7| \) in the proof?
3. How would the choice of \( \delta \) change if \( m \) were larger than 14?
4. What happens to the limit if \( x \to 7 \) instead of \( x \to 6 \)?
5. Why do we consider the minimum of \( \frac{\epsilon}{m} \) and 1 when choosing \( \delta \)?

**Tip**: In epsilon-delta proofs, always simplify and factor the function as much as possible to clearly see how \( \delta \) relates to \( \epsilon \).

In formally proving that lim_{x → 6} (x^2 + x) = 42, let ε > 0 be arbitrary. Choose δ = min(ε/m, 1). Determine the smallest value of m that would satisfy the proof.

Learn how to use the epsilon-delta definition to prove lim_{x → 6} (x^2 + x) = 42. This step-by-step guide explains how to determine the smallest value of m in δ = min(ε/m, 1) to satisfy the proof, with bounds on the algebraic expression.

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