The mathematical expression in the image is:

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

This is an indeterminate form $\frac{0}{0}$, so we need to simplify the expression by factoring the numerator.

Steps:

1. Factor $x^3 - 8$: This is a difference of cubes. We use the formula:

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

In our case, $a = x$ and $b = 2$, so:

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

2. Rewrite the limit:

$\lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{x - 2}$

3. Cancel out the common term $(x - 2)$ from the numerator and the denominator:

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

4. Substitute $x = 2$ into the remaining expression:

$2^2 + 2(2) + 4 = 4 + 4 + 4 = 12$

Final Answer:

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

Would you like more details on the steps or have any other questions?

Here are five related questions to expand on this:

1. What are the rules for evaluating limits involving indeterminate forms?
2. How can you factor the sum of cubes $x^3 + 8$?
3. What is L'Hopital's Rule, and how could it be applied to this type of problem?
4. How would you handle limits that involve more complex factorizations?
5. How can you determine when a limit does not exist?

Tip: Always check for factoring possibilities when encountering $\frac{0}{0}$ forms in limits.