Evaluating a 0/0 limit by rationalizing both radicals

Multiply the numerator by $\frac{\sqrt{9+x}+3}{\sqrt{9+x}+3}$ and the denominator by $\frac{\sqrt{4+x}+2}{\sqrt{4+x}+2}$. This uses the difference of squares identity to turn $\sqrt{9+x}-3$ into $\frac{(9+x)-9}{\sqrt{9+x}+3} = \frac{x}{\sqrt{9+x}+3}$, and similarly $\sqrt{4+x}-2$ becomes $\frac{x}{\sqrt{4+x}+2}$.

The original expression becomes $\frac{x/(\sqrt{9+x}+3)}{x/(\sqrt{4+x}+2)} = \frac{\sqrt{4+x}+2}{\sqrt{9+x}+3},$ after canceling the common factor of $x$ (valid since $x \neq 0$ near the limit point). Now the expression is continuous at $x=0$, so direct substitution gives $\frac{\sqrt{4}+2}{\sqrt{9}+3} = \frac{2+2}{3+3} = \frac{4}{6} = \frac{2}{3}$.

The common factor canceled was x. After multiplying numerator and denominator by their respective conjugates, both the top and bottom simplified to expressions containing a factor of x, which canceled since x is not zero near the limit point. This left a continuous expression that could be evaluated by direct substitution.