Using the definition of functional limit to prove that the limit as x goes to c of x^3=8. My solution is as follows: Let epsilon>0, Let delta=(epsilon/7)>0 for any 0<|x-2|<delta. so, |f(x)-L|=|x^3-8|=|(x-2)(x^2+2x+4|=|x-2| times 7 < 7 times delta = 7 times (epsilon/7)=epsilon

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 \( \lim_{x \to 2} x^3 = 8 \).

---

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!

Learn how to prove the limit of x^3 as x approaches 2 using the epsilon-delta definition of limits. This step-by-step solution covers key calculus concepts like functional limits, bounding expressions, and choosing appropriate delta values. Suitable for Grades 11-12 and college-level calculus students.

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