Your proof is almost correct but could benefit from a clearer structure, particularly in the relationship between $\delta$ and $\epsilon$. I'll guide you through a more formal approach to prove that:

$\lim_{x \to 2} x^3 = 8$

Goal:

Using the definition of the limit, we want to prove that for every $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - 2| < \delta$, then $|x^3 - 8| < \epsilon$.

Proof:

Let $\epsilon > 0$ be given. We need to find a $\delta > 0$ such that if $0 < |x - 2| < \delta$, it implies that $|x^3 - 8| < \epsilon$.

1. Start with the expression $|x^3 - 8|$: $|x^3 - 8| = |x^3 - 2^3| = |(x - 2)(x^2 + 2x + 4)|$

2. We want to make $|x^3 - 8| < \epsilon$. To achieve this, notice that: $|x^3 - 8| = |x - 2| \cdot |x^2 + 2x + 4|$

3. Control the term $|x^2 + 2x + 4|$:
   Observe that near $x = 2$, $x^2 + 2x + 4$ doesn't vary too much. For example, if $x$ is close to 2, say $1 \leq x \leq 3$, we can bound $|x^2 + 2x + 4|$ by a constant.

   • When $x = 1$, $x^2 + 2x + 4 = 1^2 + 2(1) + 4 = 7$.
   • When $x = 3$, $x^2 + 2x + 4 = 3^2 + 2(3) + 4 = 19$.

   So for $1 \leq x \leq 3$, we know: $7 \leq |x^2 + 2x + 4| \leq 19$

4. Choose $\delta$:
   To ensure $|x^3 - 8| < \epsilon$, we need: $|x - 2| \cdot |x^2 + 2x + 4| < \epsilon$ Using the fact that $|x^2 + 2x + 4| \leq 19$ for $x$ near 2, we can simplify this to: $|x - 2| \cdot 19 < \epsilon$ Therefore, choose: $\delta = \frac{\epsilon}{19}$

5. Conclusion: If $0 < |x - 2| < \delta = \frac{\epsilon}{19}$, then: $|x^3 - 8| = |x - 2| \cdot |x^2 + 2x + 4| \leq |x - 2| \cdot 19 < \frac{\epsilon}{19} \cdot 19 = \epsilon$

Thus, by the definition of the limit, we have shown that $\lim_{x \to 2} x^3 = 8$.

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Would you like more details on any specific step or further clarification?

Here are 5 related questions you might find helpful:

1. How would this proof change for a different cubic polynomial, like $x^3 + 1$?
2. Can you apply a similar method to prove the limit of $x^4$ as $x \to 2$?
3. Why is bounding $|x^2 + 2x + 4|$ important in this proof?
4. How would the limit change if we approached a different value for $c$ instead of 2?
5. What would happen if we allowed $x \to \infty$ instead of $x \to 2$?

Tip: Always look for ways to bound complicated expressions when working with limits, as this simplifies finding appropriate $\delta$-values!