To evaluate the limit of the function $f(x) = x^3 + 2x^2 - 3x - 4$ as $x$ approaches $-1$, we simply substitute $x = -1$ into the expression:

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

Substituting $x = -1$:

$f(-1) = (-1)^3 + 2(-1)^2 - 3(-1) - 4$

Now, let's calculate each term:

$(-1)^3 = -1, \quad 2(-1)^2 = 2 \times 1 = 2, \quad -3(-1) = 3, \quad -4 = -4$

Adding these together:

$f(-1) = -1 + 2 + 3 - 4 = 0$

Therefore, the limit is:

$\lim_{x \to -1} (x^3 + 2x^2 - 3x - 4) = 0$

Would you like further clarification or details on how this works?

Here are five related questions to explore further:

1. How do you find limits using L'Hôpital's Rule?
2. What happens when a limit leads to an indeterminate form like $\frac{0}{0}$?
3. Can all polynomials be solved using direct substitution for limits?
4. How do limits behave at infinity for polynomial functions?
5. What is the significance of one-sided limits?

Tip: When calculating limits of polynomials, direct substitution is usually the most efficient method unless the expression results in an indeterminate form.