Evaluate a limit by rationalizing

Multiply the numerator and denominator by the conjugate $\sqrt{x}+2$: $\frac{x-4}{\sqrt{x}-2}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2} = \frac{(x-4)(\sqrt{x}+2)}{x-4}.$ The denominator becomes $(\sqrt{x}-2)(\sqrt{x}+2) = x - 4$, matching the factor $(x-4)$ already in the numerator.

Since $x \neq 4$ as $x \to 4$, the factor $(x-4)$ cancels, leaving $\sqrt{x}+2$. Now direct substitution works: $\lim_{x\to 4}(\sqrt{x}+2) = \sqrt{4}+2 = 4$.

The answer is square root of x plus two. This is the conjugate of square root of x minus two, and multiplying by it eliminates the square root from the denominator, turning it into x minus four so the common factor cancels.