The given expression is:

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

Let's solve this step by step.

Step 1: Substitute $x = 2$ directly into the expression.

If we substitute $x = 2$ directly into the expression:

$\frac{(2^3 - 2 \times 2^2)}{(2^2 - 4)} = \frac{(8 - 8)}{(4 - 4)} = \frac{0}{0}$

We get the indeterminate form $\frac{0}{0}$. This means we need to apply algebraic simplification or use L'Hôpital's Rule.

Step 2: Factor the denominator and numerator.

Factor the denominator first:

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

Next, factor the numerator:

$x^3 - 2x^2 = x^2(x - 2)$

Step 3: Simplify the expression.

Now rewrite the limit with the factored terms:

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

We can cancel out the common factor $(x - 2)$ from both the numerator and the denominator:

$\lim_{{x \to 2}} \frac{x^2}{x + 2}$

Step 4: Substitute $x = 2$ into the simplified expression.

Now that we have eliminated the indeterminate form, we substitute $x = 2$ into the simplified expression:

$\frac{2^2}{2 + 2} = \frac{4}{4} = 1$

Final Answer:

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

Would you like further details or explanations?

Here are 5 related questions to expand on this concept:

1. What is L'Hôpital's Rule, and when is it used to solve limits?
2. How do you factor polynomials like $x^3 - 2x^2$?
3. What other techniques can be used to simplify indeterminate forms like $\frac{0}{0}$?
4. Can limits involving higher-order polynomials also be simplified by factoring?
5. How does the behavior of limits help in understanding continuity of functions?

Tip: Always check for the possibility of factoring polynomials when you encounter indeterminate forms like $\frac{0}{0}$ in limits.