The given limit is:

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

We can attempt to solve this by substituting $x = 3$ directly first:

$\frac{3^2 - 2(3) - 3}{3 - 3} = \frac{9 - 6 - 3}{0} = \frac{0}{0}$

This results in an indeterminate form $\frac{0}{0}$, so we need to simplify the expression further. Let's factor the numerator:

$x^2 - 2x - 3 = (x - 3)(x + 1)$

Thus, the expression becomes:

$\frac{(x - 3)(x + 1)}{x - 3}$

Now, cancel out the $(x - 3)$ terms:

$\lim_{x \to 3} (x + 1)$

Now substitute $x = 3$ into this simplified expression:

$3 + 1 = 4$

So the value of the limit is:

$\boxed{4}$

Would you like further details on this solution or have any questions?

Here are five related questions you may find helpful:

1. What is the difference between the indeterminate form $\frac{0}{0}$ and a defined value in limits?
2. Can we apply this factoring method to all types of limits with indeterminate forms?
3. How can we use L'Hopital's Rule for limits that result in $\frac{0}{0}$?
4. What is the behavior of a function near a point of discontinuity?
5. How do we interpret limits approaching infinity in rational functions?

Tip: Always look for ways to simplify the expression, like factoring, before attempting to evaluate the limit.