Evaluate a limit by rationalizing the numerator

Multiply the numerator and denominator by the conjugate $\sqrt{x}+2$: $\frac{\sqrt{x}-2}{x-4}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2} = \frac{x-4}{(x-4)(\sqrt{x}+2)}.$ This eliminates the radical from the numerator, since $(\sqrt{x}-2)(\sqrt{x}+2) = x - 4$.

For $x \ne 4$, the factor $x-4$ cancels, leaving $\frac{1}{\sqrt{x}+2}.$ This new expression is continuous at $x=4$, so direct substitution gives $\frac{1}{\sqrt{4}+2} = \frac{1}{4}$.