The problem is to evaluate the following limit:

$\lim_{x \to 2} \frac{x^3 - 8}{x - 2}.$

Solution:

1. Factorize the numerator:

   Notice that the numerator $x^3 - 8$ is a difference of cubes, which can be factorized as follows:

   $x^3 - 8 = (x - 2)(x^2 + 2x + 4).$

2. Simplify the expression:

   Substitute the factorization into the limit:

   $\frac{x^3 - 8}{x - 2} = \frac{(x - 2)(x^2 + 2x + 4)}{x - 2}.$

   Since $x \neq 2$, the $(x - 2)$ terms cancel:

   $= x^2 + 2x + 4.$

3. Evaluate the limit:

   Substitute $x = 2$ into the simplified expression:

   $\lim_{x \to 2} (x^2 + 2x + 4) = 2^2 + 2(2) + 4 = 4 + 4 + 4 = 12.$

Final Answer:

$\lim_{x \to 2} \frac{x^3 - 8}{x - 2} = 12.$

Would you like any further details or have any questions?

Related Questions:

1. How do you factorize a sum of cubes?
2. What are the common techniques to solve indeterminate forms in limits?
3. How does the difference of cubes formula work?
4. What are other ways to solve limits besides factoring?
5. Can L'Hôpital's Rule be applied to this limit?

Tip: Always check for factorization possibilities when faced with indeterminate forms like $\frac{0}{0}$.