The problem in the image deals with finding the limit of the expression $\lim_{x \to 2} \frac{(3x - 1)^3 - (x + 3)^3}{x^3 - 8}$.

Solution breakdown:

1. Direct substitution:

   • First, the solution substitutes $x = 2$ into the expression, leading to the form $\frac{0}{0}$, which is indeterminate. This indicates that a different approach is needed to evaluate the limit.

2. Factoring the numerator:

   • The solution proceeds by factoring the cubic expressions in the numerator.
   • Both $(3x - 1)^3$ and $(x + 3)^3$ are expanded and simplified.
   • After factoring out common terms, the numerator can be expressed in a factored form that cancels out the indeterminate part with the denominator.

3. Simplifying:

   • The factor $(x - 2)$ is factored out of the numerator and cancels with the corresponding factor in the denominator, leaving a simpler expression to substitute $x = 2$.

4. Final substitution:

   • After canceling and simplifying, the limit can be evaluated by direct substitution, which yields the final result $\frac{25}{2}$.

Thus, the final answer is:

$\lim_{x \to 2} \frac{(3x - 1)^3 - (x + 3)^3}{x^3 - 8} = \frac{25}{2}$

Would you like more details on any step?

Here are 5 questions to expand your understanding:

1. Why is it necessary to factor the expression when the direct substitution gives $\frac{0}{0}$?
2. How is the difference of cubes formula used in this problem?
3. What are the steps to expand $(3x - 1)^3$ and $(x + 3)^3$?
4. How do we factor out the common terms from the numerator and denominator?
5. What would happen if we used L'Hopital's Rule instead of factoring?

Tip: When facing indeterminate forms like $\frac{0}{0}$, look for factoring or simplifying techniques to resolve the limit.