To find the limit of $\frac{\sqrt[3]{x} - 2}{x - 8}$ as $x$ approaches 8, we'll follow these steps:

Step 1: Direct Substitution

First, substitute $x = 8$ directly into the expression:

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

This results in an indeterminate form $\frac{0}{0}$, so we need to use another method to evaluate the limit.

Step 2: Rationalize the Numerator

To resolve the indeterminate form, we can multiply the numerator and the denominator by the conjugate of the numerator's cube root, using the identity:

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Let $a = \sqrt[3]{x}$ and $b = 2$. Then:

[ a^3 - b^3 = x - 8 = (\sqrt[3]{x} - 2)(\sqrt[3]{x^2} + 2\sqrt[3]{x} + 4) ]

So,

[ \frac{\sqrt[3]{x} - 2}{x - 8} = \frac{(\sqrt[3]{x} - 2)(\sqrt[3]{x^2} + 2\sqrt[3]{x} + 4)}{(x - 8)(\sqrt[3]{x^2} + 2\sqrt[3]{x} + 4)} ]

Since $x - 8 = (\sqrt[3]{x} - 2)(\sqrt[3]{x^2} + 2\sqrt[3]{x} + 4)$, the expression simplifies to:

[ \frac{1}{\sqrt[3]{x^2} + 2\sqrt[3]{x} + 4} ]

Step 3: Take the Limit

Now, substitute $x = 8$ into the simplified expression:

[ \lim_{x \to 8} \frac{1}{\sqrt[3]{x^2} + 2\sqrt[3]{x} + 4} = \frac{1}{\sqrt[3]{8^2} + 2\sqrt[3]{8} + 4} = \frac{1}{\sqrt[3]{64} + 2 \times 2 + 4} = \frac{1}{4 + 4 + 4} = \frac{1}{12} ]

Final Answer

The limit is:

$\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x - 8} = \frac{1}{12}$

Would you like any details or have further questions?

Here are five related questions you might find interesting:

1. How would you evaluate $\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$?
2. Can L'Hôpital's Rule be used to solve this limit problem?
3. What is the geometric interpretation of limits in calculus?
4. How do you find the limit of a function as $x$ approaches infinity?
5. What are other methods to resolve indeterminate forms besides rationalization?

Tip: When you encounter indeterminate forms, always consider algebraic manipulation, such as rationalization or factoring, before resorting to more advanced techniques.