We are tasked with evaluating the limit:

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

Direct substitution of $x = 8$ results in an indeterminate form $\frac{0}{0}$, so we must use algebraic techniques to simplify the expression. In this case, we can use the technique of rationalizing the numerator, as the cube root function is involved.

Step 1: Rewrite the expression

We need to simplify the given limit:

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

To rationalize, we multiply both the numerator and the denominator by the expression that will eliminate the cube root in the numerator. This is:

[ \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)} ]

Step 2: Simplify using the identity

Using the identity $(a - b)(a^2 + ab + b^2) = a^3 - b^3$, where $a = \sqrt[3]{x}$ and $b = 2$, we get:

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

Thus, the expression becomes:

[ \frac{x - 8}{(x - 8)(\sqrt[3]{x^2} + 2\sqrt[3]{x} + 4)} = \frac{1}{\sqrt[3]{x^2} + 2\sqrt[3]{x} + 4} ]

Step 3: Evaluate the limit

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

[ \frac{1}{\sqrt[3]{8^2} + 2\sqrt[3]{8} + 4} = \frac{1}{\sqrt[3]{64} + 2 \cdot \sqrt[3]{8} + 4} ]

Since $\sqrt[3]{64} = 4$ and $\sqrt[3]{8} = 2$, we have:

$\frac{1}{4 + 2 \cdot 2 + 4} = \frac{1}{4 + 4 + 4} = \frac{1}{12}$

Final Answer:

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

Would you like further explanation or details on any step? Here are some related questions you can explore:

1. How can you rationalize cube roots in other types of limits?
2. What is the difference between using direct substitution versus rationalization in limits?
3. How would this process change for a square root instead of a cube root?
4. Can L'Hopital's Rule be used to solve this limit problem?
5. What happens when you apply limits at infinity for similar rational expressions?

Tip: When dealing with indeterminate forms, rationalizing can be an effective tool for simplifying expressions, especially with roots.