The given limit is:

$\lim_{x \to 2} \frac{x^3 - 2x^2}{x^2 - 4}$

Step 1: Simplify the expression

Notice that the denominator $x^2 - 4$ can be factored as: $x^2 - 4 = (x - 2)(x + 2)$

The numerator $x^3 - 2x^2$ can be factored as: $x^3 - 2x^2 = x^2(x - 2)$

Now the expression becomes: $\frac{x^3 - 2x^2}{x^2 - 4} = \frac{x^2(x - 2)}{(x - 2)(x + 2)}$

Step 2: Cancel common terms

For $x \neq 2$, we can cancel $(x - 2)$ from the numerator and denominator: $\frac{x^2(x - 2)}{(x - 2)(x + 2)} = \frac{x^2}{x + 2}, \quad x \neq 2$

Step 3: Substitute $x = 2$

Substitute $x = 2$ into the simplified expression: $\frac{x^2}{x + 2} = \frac{2^2}{2 + 2} = \frac{4}{4} = 1$

Final Answer:

$\lim_{x \to 2} \frac{x^3 - 2x^2}{x^2 - 4} = 1$

Let me know if you'd like further clarification or a detailed explanation of any step.

Here are some related questions to expand your understanding:

1. How do we handle limits involving indeterminate forms such as $\frac{0}{0}$?
2. What are other techniques for solving limits besides factoring?
3. How does L'Hôpital's Rule apply to this problem?
4. Can this limit be solved using numerical approximation? How?
5. What happens if $x \to -2$ instead of $x \to 2$?

Tip: Always check for factorizations or common terms before applying more advanced limit-solving techniques like L'Hôpital's Rule.