To find the limit:

$\lim_{x \to -2} \frac{x^3 + 8}{x^2 - 4}$

Step 1: Factor both the numerator and the denominator.

The numerator, $x^3 + 8$, is a sum of cubes, which can be factored as:

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

The denominator, $x^2 - 4$, is a difference of squares and can be factored as:

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

Step 2: Simplify the expression.

Now, the expression becomes:

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

You can cancel out the common factor of $(x + 2)$, provided $x \neq -2$. This simplifies to:

$\frac{x^2 - 2x + 4}{x - 2}$

Step 3: Evaluate the limit as $x \to -2$.

Now substitute $x = -2$ into the simplified expression:

$\frac{(-2)^2 - 2(-2) + 4}{-2 - 2} = \frac{4 + 4 + 4}{-4} = \frac{12}{-4} = -3$

Final Answer:

$\lim_{x \to -2} \frac{x^3 + 8}{x^2 - 4} = -3$

Would you like further details or clarification on any of the steps?

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Here are 5 related questions you might find interesting:

1. How would the limit change if the expression was $\frac{x^3 - 8}{x^2 - 4}$?
2. What happens when we approach a limit that results in a $\frac{0}{0}$ indeterminate form?
3. Can you explain the significance of canceling common factors when simplifying limits?
4. How does the limit differ when approaching from the right ($x \to -2^+$) vs. from the left ($x \to -2^-$)?
5. What would happen if we tried to evaluate $\lim_{x \to 2} \frac{x^3 + 8}{x^2 - 4}$?

Tip: Always check for possible factorizations when encountering limits with polynomials to simplify indeterminate forms.