Evaluating a limit by rationalizing the numerator

Multiply numerator and denominator by the conjugate: $\frac{\sqrt{x+9}-3}{x} \cdot \frac{\sqrt{x+9}+3}{\sqrt{x+9}+3} = \frac{(x+9) - 9}{x(\sqrt{x+9}+3)} = \frac{x}{x(\sqrt{x+9}+3)}.$ This removes the square root from the numerator, turning it into a difference of squares.

Cancel the common factor of $x$ (valid since $x \to 0$ means $x \neq 0$) to get $\frac{1}{\sqrt{x+9}+3}.$ This new expression is continuous at $x=0$, so by the quotient rule for limits we substitute directly: $\sqrt{0+9}+3 = 6$, giving a limit of $1/6$.