The given limit is:

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

First, simplify the numerator:

[ (3 + x)^2 - 3^2 = [(3 + x) - 3][(3 + x) + 3] ] This follows from the difference of squares formula, $a^2 - b^2 = (a - b)(a + b)$.

Now, apply this:

$(3 + x)^2 - 3^2 = (x)(x + 6)$

Thus, the expression becomes:

$\lim_{x \to 3} \frac{x(x + 6)}{x - 3}$

Since direct substitution leads to $0/0$, we can factor and cancel terms. Rewrite the expression:

$\lim_{x \to 3} \frac{x(x + 6)}{x - 3}$

This cannot be simplified directly. Instead, let's re-examine the structure or use L'Hopital's Rule, if necessary.

Do you need further details or have questions?

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Questions for exploration:

1. How is L'Hopital's Rule used in limit problems?
2. What is the difference of squares formula?
3. How do you factor higher-degree polynomials?
4. How can we handle indeterminate forms in limits?
5. How does substitution work in limits?

Tip: When encountering $0/0$ forms, always consider factoring or applying L'Hopital’s Rule.