Evaluate a 0/0 limit by rationalizing the numerator

Direct substitution at $x=4$ gives $\frac{\sqrt{4}-2}{4-4} = \frac{0}{0}$, an indeterminate form, so we need an algebraic technique before applying the limit laws. Multiply the numerator and denominator by the conjugate $\sqrt{x}+2$: $\frac{\sqrt{x}-2}{x-4}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2} = \frac{x-4}{(x-4)(\sqrt{x}+2)}.$

For $x \neq 4$, the factor $(x-4)$ cancels, leaving $\frac{1}{\sqrt{x}+2}$, which is continuous at $x=4$. Now applying the quotient rule and direct substitution: $\lim_{x\to 4} \frac{1}{\sqrt{x}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{4}.$

The answer is square root of x plus two. After multiplying by the conjugate and canceling the common factor of x minus four, the expression simplified to one divided by the square root of x plus two. This new expression is continuous at four, so direct substitution could be applied to find the final answer of one fourth.