To solve this limit problem using L'Hôpital's Rule, let's go through it step-by-step.

The problem is to find: $\lim_{{x \to 8}} \frac{x^3 - 63x - 8}{x^3 - 512}$

Step 1: Check the Form of the Limit

1. Substitute $x = 8$ into the expression:
   • Numerator: $8^3 - 63 \cdot 8 - 8 = 512 - 504 - 8 = 0$
   • Denominator: $8^3 - 512 = 512 - 512 = 0$

Since substituting $x = 8$ results in $\frac{0}{0}$, we can apply L'Hôpital's Rule, which requires us to differentiate the numerator and denominator separately.

Step 2: Differentiate the Numerator and the Denominator

1. Numerator: $f(x) = x^3 - 63x - 8$

   • $f'(x) = 3x^2 - 63$

2. Denominator: $g(x) = x^3 - 512$

   • $g'(x) = 3x^2$

Step 3: Apply L'Hôpital's Rule

Now we find the limit of the new expression: $\lim_{{x \to 8}} \frac{f'(x)}{g'(x)} = \lim_{{x \to 8}} \frac{3x^2 - 63}{3x^2}$

This can be simplified by canceling out the common factor of 3 in the numerator and denominator: $= \lim_{{x \to 8}} \frac{x^2 - 21}{x^2}$

Step 4: Simplify the Expression

Rewrite the expression as: $= \lim_{{x \to 8}} \left(1 - \frac{21}{x^2}\right)$

Step 5: Substitute $x = 8$ into the Simplified Expression

$= 1 - \frac{21}{8^2} = 1 - \frac{21}{64} = \frac{64 - 21}{64} = \frac{43}{64}$

Final Answer

$\lim_{{x \to 8}} \frac{x^3 - 63x - 8}{x^3 - 512} = \frac{43}{64}$

Would you like more details on any of these steps, or do you have any questions?

Here are five additional questions to expand on this topic:

1. How does L'Hôpital's Rule apply to indeterminate forms like $\frac{0}{0}$?
2. What other types of indeterminate forms can be resolved using L'Hôpital's Rule?
3. Can we use L'Hôpital's Rule more than once on the same problem?
4. How would we solve this limit if L'Hôpital’s Rule wasn’t allowed?
5. What is the significance of simplifying the expression before substituting values?

Tip: Always check if a limit is in an indeterminate form before applying L'Hôpital's Rule. This ensures the method is valid for that limit problem.