The limit expression given is:

$\lim_{x \to 4} \sqrt{x^2 - 2x - 4}$

Step-by-Step Solution:

1. Check if the function is continuous:

   • The function involves a square root, so it's important to check that the expression inside the square root, $x^2 - 2x - 4$, is non-negative at and near $x = 4$, because the square root function is only defined for non-negative values in the reals.

2. Substitute $x = 4$ directly into the function: Since the limit involves a continuous function (square root), we can try direct substitution if the expression inside the square root is non-negative.

   First, evaluate $x^2 - 2x - 4$ at $x = 4$:

   $4^2 - 2(4) - 4 = 16 - 8 - 4 = 4$

   Thus, the expression inside the square root becomes:

   $\sqrt{4} = 2$

3. Conclusion: Since substituting $x = 4$ into the expression gives a real and finite result, the limit is:

   $\boxed{2}$

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

Here are five related questions to deepen your understanding:

1. How do you determine if a function involving a square root is continuous?
2. What would happen if the expression inside the square root were negative at $x = 4$?
3. How would you approach solving this limit if direct substitution didn't work?
4. What is the general form of a continuous function, and why can we directly substitute in this case?
5. How can you apply the limit laws to solve more complex limits involving square roots?

Tip: Always check the domain of functions involving square roots or logarithms before applying limit rules directly!