The problem asks to find the value of the following limit:

$\lim_{x \to 4} \frac{2x - 8}{\sqrt{x} - 2}$

Step-by-step Solution:

1. Direct Substitution: Let's first try substituting $x = 4$ directly into the expression.

   Numerator: $2(4) - 8 = 8 - 8 = 0$

   Denominator: $\sqrt{4} - 2 = 2 - 2 = 0$

   Since both the numerator and denominator become 0, we have the indeterminate form $\frac{0}{0}$. Therefore, we need to simplify the expression further.

2. Factoring the expression: Let's simplify by manipulating the numerator and denominator.

   The numerator $2x - 8$ can be factored as: $2x - 8 = 2(x - 4)$

   Now the expression becomes: $\frac{2(x - 4)}{\sqrt{x} - 2}$

3. Multiplying by the conjugate: To eliminate the square root in the denominator, multiply both the numerator and denominator by the conjugate of the denominator, $\sqrt{x} + 2$: $\frac{2(x - 4)}{\sqrt{x} - 2} \times \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \frac{2(x - 4)(\sqrt{x} + 2)}{(\sqrt{x} - 2)(\sqrt{x} + 2)}$

   Using the difference of squares in the denominator: $(\sqrt{x})^2 - 2^2 = x - 4$

   The expression simplifies to: $\frac{2(x - 4)(\sqrt{x} + 2)}{x - 4}$

4. Cancel common terms: Since $x - 4$ appears in both the numerator and denominator, we can cancel them (assuming $x \neq 4$): $2(\sqrt{x} + 2)$

5. Substitute $x = 4$: Now, substitute $x = 4$ into the simplified expression: $2(\sqrt{4} + 2) = 2(2 + 2) = 2(4) = 8$

Thus, the value of the limit is:

$\boxed{8}$

Let me know if you'd like more details or have any questions!

Here are 5 additional questions to deepen your understanding:

1. What is the difference of squares and why does it help simplify expressions with square roots?
2. Can you explain why the limit resulted in an indeterminate form initially?
3. How would the limit change if the denominator had a different expression, like $\sqrt{x} - 1$?
4. Could L'Hopital's Rule have been used to solve this limit problem? If so, how?
5. How do you factor out terms when simplifying expressions involving limits?

Tip: Always try direct substitution first in limit problems. If you encounter an indeterminate form, look for algebraic techniques like factoring or rationalizing!