1. **State the problem:** Find the limit as $x$ approaches 0 of the expression $$\frac{\sqrt{x + 9} - 3}{\sqrt{x + 16} - 4}.$$\n\n2. **Recall the formula and approach:** When direct substitution leads to an indeterminate form like $\frac{0}{0}$, we use algebraic manipulation such as rationalizing the numerator or denominator to simplify the expression.\n\n3. **Check direct substitution:** Substitute $x=0$:\n$$\frac{\sqrt{0 + 9} - 3}{\sqrt{0 + 16} - 4} = \frac{3 - 3}{4 - 4} = \frac{0}{0},$$ which is indeterminate.\n\n4. **Rationalize numerator and denominator:** Multiply numerator and denominator by the conjugates to simplify:\n$$\frac{\sqrt{x + 9} - 3}{\sqrt{x + 16} - 4} \times \frac{\sqrt{x + 9} + 3}{\sqrt{x + 9} + 3} \times \frac{\sqrt{x + 16} + 4}{\sqrt{x + 16} + 4}.$$\n\n5. **Simplify numerator:**\n$$ (\sqrt{x + 9} - 3)(\sqrt{x + 9} + 3) = (x + 9) - 9 = x.$$\n\n6. **Simplify denominator:**\n$$ (\sqrt{x + 16} - 4)(\sqrt{x + 16} + 4) = (x + 16) - 16 = x.$$\n\n7. **Rewrite the expression:**\n$$\frac{x}{x} \times \frac{\sqrt{x + 16} + 4}{\sqrt{x + 9} + 3} = \cancel{\frac{x}{x}} \times \frac{\sqrt{x + 16} + 4}{\sqrt{x + 9} + 3}.$$\n\n8. **Cancel common factor $x$:**\n$$1 \times \frac{\sqrt{x + 16} + 4}{\sqrt{x + 9} + 3} = \frac{\sqrt{x + 16} + 4}{\sqrt{x + 9} + 3}.$$\n\n9. **Evaluate the limit by direct substitution now:**\n$$\lim_{x \to 0} \frac{\sqrt{x + 16} + 4}{\sqrt{x + 9} + 3} = \frac{\sqrt{0 + 16} + 4}{\sqrt{0 + 9} + 3} = \frac{4 + 4}{3 + 3} = \frac{8}{6} = \frac{4}{3}.$$\n\n**Final answer:** $$\boxed{\frac{4}{3}}.$$