Evaluating a limit by rationalizing the numerator

Substituting $x=4$ gives $\frac{\sqrt{9}-3}{0} = \frac{0}{0}$, an indeterminate form, so direct substitution fails and we need an algebraic technique. Since the numerator contains a square root, multiply numerator and denominator by the conjugate $\sqrt{x+5}+3$: $\frac{\sqrt{x+5}-3}{x-4}\cdot\frac{\sqrt{x+5}+3}{\sqrt{x+5}+3} = \frac{(x+5)-9}{(x-4)(\sqrt{x+5}+3)}.$

The numerator simplifies using the difference of squares formula $(\sqrt{x+5})^2 - 3^2 = (x+5) - 9 = x-4$, so the expression becomes $\frac{x-4}{(x-4)(\sqrt{x+5}+3)}.$ Since $x \ne 4$ near the limit, cancel the common factor $x-4$ to get $\frac{1}{\sqrt{x+5}+3}$. Now substitute $x=4$: $\frac{1}{\sqrt{9}+3} = \frac{1}{6}$.

The conjugate used was the square root of x plus five, plus three. This expression was multiplied by both the numerator and denominator so that squaring the square root term would eliminate the radical, turning the numerator into a difference that matches the denominator and allows cancellation.