We are asked to evaluate the following limit:

$\lim_{x \to -3} \frac{x + 3}{x^3 + 27}$

First, observe that the denominator can be factored using the sum of cubes formula:

$x^3 + 27 = (x + 3)(x^2 - 3x + 9)$

Thus, the original expression becomes:

$\frac{x + 3}{(x + 3)(x^2 - 3x + 9)}$

For $x \neq -3$, we can cancel the $x + 3$ term in the numerator and denominator:

$\frac{1}{x^2 - 3x + 9}$

Now, we can directly substitute $x = -3$ into this simplified expression:

$\frac{1}{(-3)^2 - 3(-3) + 9} = \frac{1}{9 + 9 + 9} = \frac{1}{27}$

Therefore, the limit is:

$\boxed{\frac{1}{27}}$

Would you like a deeper explanation or have any other questions?

Here are 5 related questions:

1. What is the sum of cubes formula?
2. How would the limit change if the expression was $x^3 - 27$?
3. What happens if we try to evaluate the original limit without factoring?
4. How can you apply L'Hopital's Rule to this limit?
5. What are the conditions under which limits of rational functions are undefined?

Tip: Always look for factoring techniques like sum or difference of cubes to simplify expressions when evaluating limits.