The expression provided is:

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

To solve this limit, let's first substitute $x = 4$ directly into the expression:

$\frac{{4 - 4}}{{\sqrt{4} - 2}} = \frac{0}{0}$

This results in an indeterminate form (0/0), so we need to simplify the expression. A common technique here is to multiply by the conjugate of the denominator. Let's rewrite the expression and multiply by the conjugate of $\sqrt{x} - 2$, which is $\sqrt{x} + 2$:

$\lim_{{x \to 4}} \frac{{x - 4}}{{\sqrt{x} - 2}} \cdot \frac{{\sqrt{x} + 2}}{{\sqrt{x} + 2}} = \lim_{{x \to 4}} \frac{{(x - 4)(\sqrt{x} + 2)}}{{(\sqrt{x} - 2)(\sqrt{x} + 2)}}$

The denominator becomes:

$(\sqrt{x} - 2)(\sqrt{x} + 2) = x - 4$

Thus, the expression simplifies to:

$\lim_{{x \to 4}} \frac{{(x - 4)(\sqrt{x} + 2)}}{{x - 4}} = \lim_{{x \to 4}} (\sqrt{x} + 2)$

Since the $x - 4$ terms cancel out, we are left with:

$\lim_{{x \to 4}} (\sqrt{x} + 2)$

Now substitute $x = 4$ into the remaining expression:

$\sqrt{4} + 2 = 2 + 2 = 4$

Therefore, the value of the limit is:

$\boxed{4}$

Would you like more details on any of the steps? Here are some related questions you might consider:

1. What other methods can be used to solve limits with indeterminate forms?
2. How does multiplying by the conjugate simplify expressions involving square roots?
3. What are other examples of limits that result in the indeterminate form $\frac{0}{0}$?
4. Can L'Hôpital's Rule be used to solve this type of limit? If so, how?
5. How does the concept of a limit apply to real-world scenarios?

Tip: When encountering the indeterminate form $\frac{0}{0}$, consider simplifying the expression using algebraic techniques like factoring, multiplying by conjugates, or applying L'Hôpital's Rule.