1. **State the problem:** Find the limit $$\lim_{x \to 4} \frac{x-4}{\sqrt{x}-2}$$. 2. **Recall the formula and rules:** Direct substitution gives $$\frac{4-4}{\sqrt{4}-2} = \frac{0}{0}$$ which is an indeterminate form. We need to simplify the expression. 3. **Simplify the expression:** Multiply numerator and denominator by the conjugate of the denominator: $$\frac{x-4}{\sqrt{x}-2} \cdot \frac{\sqrt{x}+2}{\sqrt{x}+2} = \frac{(x-4)(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}$$ 4. **Use the difference of squares in the denominator:** $$ (\sqrt{x}-2)(\sqrt{x}+2) = x - 4 $$ 5. **Substitute back:** $$ \frac{(x-4)(\sqrt{x}+2)}{x-4} $$ 6. **Cancel common factor:** $$ \frac{\cancel{x-4}(\sqrt{x}+2)}{\cancel{x-4}} = \sqrt{x} + 2 $$ 7. **Evaluate the limit by direct substitution:** $$ \lim_{x \to 4} (\sqrt{x} + 2) = \sqrt{4} + 2 = 2 + 2 = 4 $$ **Final answer:** $$4$$