prove the statement using the precise definition of a limit lim(x^2+2x-15)=9

To prove the limit \(\lim_{x \to 3} (x^2 + 2x - 15) = 9\) using the precise (ε-δ) definition of a limit, we must show that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that whenever \(0 < |x - 3| < \delta\), we have:

\[
|(x^2 + 2x - 15) - 9| < \epsilon
\]

### Step-by-Step Proof:

1. **Restate the expression we need to bound:**
   We are tasked with bounding:

   \[
   |(x^2 + 2x - 15) - 9| = |x^2 + 2x - 24|
   \]

   This simplifies to:

   \[
   |x^2 + 2x - 24| = |(x - 3)(x + 8)|
   \]

2. **Find a relationship between \(\delta\) and \(\epsilon\):**
   We want to find \(\delta\) such that whenever \(0 < |x - 3| < \delta\), we have:

   \[
   |(x - 3)(x + 8)| < \epsilon
   \]

3. **Bound \(x + 8\):**
   To control \(x + 8\), note that when \(x\) is close to 3, \(x + 8\) is close to 11. So, we can limit the range of \(x\) by assuming \(|x - 3| < 1\). This implies:

   \[
   2 < x < 4 \quad \text{and thus} \quad 10 < x + 8 < 12
   \]

   Therefore, \( |x + 8| \leq 12 \) for \( |x - 3| < 1 \).

4. **Control the product \(|(x - 3)(x + 8)|\):**
   Now, we can use this bound on \( |x + 8| \) to simplify the inequality:

   \[
   |(x - 3)(x + 8)| \leq 12|x - 3|
   \]

   To satisfy the condition \(|(x - 3)(x + 8)| < \epsilon\), it suffices to choose \(\delta\) such that:

   \[
   12|x - 3| < \epsilon
   \]

5. **Choose \(\delta\):**
   Solving for \(|x - 3|\), we get:

   \[
   |x - 3| < \frac{\epsilon}{12}
   \]

   Therefore, we can choose \(\delta = \min\left(1, \frac{\epsilon}{12}\right)\). This ensures that both \(|x - 3| < \delta\) and \(|x + 8|\) remains bounded.

### Conclusion:

By choosing \(\delta = \min\left(1, \frac{\epsilon}{12}\right)\), we have shown that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that whenever \(0 < |x - 3| < \delta\), it follows that:

\[
|(x^2 + 2x - 15) - 9| < \epsilon
\]

Thus, \(\lim_{x \to 3} (x^2 + 2x - 15) = 9\), as required by the precise definition of a limit.

---

Would you like further details or have any questions?

Here are 5 related questions for practice:
1. Prove \(\lim_{x \to 2} (x^2 + 3x - 4) = 6\) using the \(\epsilon\)-\(\delta\) definition.
2. Show that \(\lim_{x \to 1} (x^3 - 1) = 0\) using the precise definition of a limit.
3. Can you use the \(\epsilon\)-\(\delta\) definition to prove \(\lim_{x \to 0} x^2 = 0\)?
4. Prove \(\lim_{x \to 5} (3x - 2) = 13\) using the \(\epsilon\)-\(\delta\) definition.
5. Use the \(\epsilon\)-\(\delta\) definition to prove \(\lim_{x \to 4} \frac{1}{x} = \frac{1}{4}\).

**Tip:** Always simplify the expression inside the limit as much as possible before applying the \(\epsilon\)-\(\delta\) definition. This can make the calculations easier.

Prove the statement using the precise definition of a limit: lim(x^2 + 2x - 15) = 9

Learn how to prove the limit of the quadratic expression x^2 + 2x - 15 equals 9 using the precise epsilon-delta definition. This problem covers key calculus concepts and is suitable for high school and early college students.

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