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.