https://file.aiquickdraw.com/mathbot/file/17251738343286r7qvwj2.jpg

Let's go through each limit step by step.

### Problem (a)
\[
\lim_{x \to 1} \frac{\sqrt[3]{x + 2} - \sqrt[3]{3}}{x - 1}
\]

1. **Simplify the expression:**

   The expression involves the difference of cube roots. We can use the identity for the difference of cubes to simplify:
   \[
   \sqrt[3]{x + 2} - \sqrt[3]{3} = \frac{(x + 2) - 3}{\sqrt[3]{(x + 2)^2} + \sqrt[3]{(x + 2) \cdot 3} + \sqrt[3]{3^2}}
   \]

   However, directly applying this might be cumbersome. Instead, we can try using L'Hôpital's Rule since this is an indeterminate form \( \frac{0}{0} \):

2. **Apply L'Hôpital's Rule:**

   Differentiate the numerator and denominator with respect to \(x\):

   - The derivative of the numerator:
     \[
     \text{If } f(x) = \sqrt[3]{x + 2}, \text{ then } f'(x) = \frac{1}{3}\cdot \frac{1}{\sqrt[3]{(x + 2)^2}}
     \]

   - The derivative of the denominator:
     \[
     \text{Derivative of } x - 1 \text{ is } 1
     \]

   So applying L'Hôpital's Rule:
   \[
   \lim_{x \to 1} \frac{\frac{1}{3} \cdot \frac{1}{\sqrt[3]{(x + 2)^2}}}{1} = \frac{1}{3} \cdot \frac{1}{\sqrt[3]{3^2}}
   \]

   Substitute \( x = 1 \):
   \[
   \frac{1}{3 \times \sqrt[3]{9}} = \frac{1}{3 \times 2.08008} \approx \frac{1}{6.24024}
   \]

   Therefore, the limit is approximately:
   \[
   \lim_{x \to 1} \frac{\sqrt[3]{x + 2} - \sqrt[3]{3}}{x - 1} \approx 0.1603
   \]

### Problem (b)
This limit was already solved in the previous interaction:
\[
\lim_{y \to 2} \frac{|2y - 4|}{4y - 8 - y^3 + 2y^2} = -\frac{1}{2}
\]

### Problem (c)
This limit was also solved earlier:
\[
\lim_{z \to -\infty} \frac{\sqrt{9z^2 - 1}}{z + 3} = -3
\]

---

Would you like to explore more details or have any specific questions about these solutions?

Here are some related questions to consider:
1. How do you decide when to apply L'Hôpital's Rule to evaluate a limit?
2. What are some techniques for simplifying limits involving cube roots or other radicals?
3. What other methods can be used if L'Hôpital's Rule is not applicable?
4. How does the behavior of the function near the point of interest affect the limit?
5. What role do indeterminate forms play in the process of finding limits?

**Tip:** When working with limits that involve radicals, sometimes rationalizing or using identities like the difference of cubes or squares can simplify the process.

Explore how to evaluate limits involving cube roots and apply L'Hôpital's Rule effectively. Includes detailed solutions for limits as x approaches 1, 2, and -∞. Suitable for advanced high school students and those preparing for calculus exams.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation