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.

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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.