https://file.aiquickdraw.com/mathbot/file/1725889952737lzfdrugs.jpg

The given mathematical expression is a limit problem, and it appears to be:

\[
\lim_{{x \to 8}} \frac{\sqrt[3]{x} - 2}{x - 8}
\]

### Steps to solve:

1. **Direct Substitution**: First, attempt to substitute \( x = 8 \) directly into the expression.

\[
\frac{\sqrt[3]{8} - 2}{8 - 8} = \frac{2 - 2}{0} = \frac{0}{0}
\]

Since direct substitution results in the indeterminate form \( \frac{0}{0} \), we need to apply **L'Hôpital's Rule** or use algebraic manipulation to resolve this.

2. **L'Hôpital's Rule**: This rule states that for limits in the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can differentiate the numerator and denominator separately.

- **Numerator**: The derivative of \( \sqrt[3]{x} \) is:

\[
\frac{d}{dx} \left( \sqrt[3]{x} \right) = \frac{1}{3} x^{-2/3}
\]

- **Denominator**: The derivative of \( x - 8 \) is simply \( 1 \).

Now applying L'Hôpital's Rule:

\[
\lim_{{x \to 8}} \frac{\frac{1}{3} x^{-2/3}}{1} = \frac{1}{3} \cdot 8^{-2/3} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}
\]

### Final Answer:
\[
\lim_{{x \to 8}} \frac{\sqrt[3]{x} - 2}{x - 8} = \frac{1}{12}
\]

Would you like a more detailed breakdown, or do you have any questions? 

Here are five related questions to expand on this:

1. How does L'Hôpital's Rule help in evaluating limits of indeterminate forms?
2. What is the general process for finding the derivative of a cube root function?
3. Why is direct substitution important in limit problems?
4. What other algebraic techniques can be used to avoid L'Hôpital’s Rule?
5. How can you determine whether to apply L'Hôpital's Rule or algebraic manipulation?

**Tip**: Always check if direct substitution gives a defined value before applying L'Hôpital's Rule or other advanced techniques.

Evaluate the limit as x approaches 8 for the expression (cube root of x - 2) divided by (x - 8).

Solve the limit of the expression (cube root of x - 2) divided by (x - 8) as x approaches 8 using L'Hôpital's Rule. This problem involves evaluating indeterminate forms and differentiating cube root functions, suitable for college-level calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation