1. **State the problem:** Find the limit as $x$ approaches 3 of the expression $$\frac{\sqrt{x+13} - 2\sqrt{x+1}}{x^2 - 9}.$$\n\n2. **Recall the formula and rules:** The denominator can be factored as a difference of squares: $$x^2 - 9 = (x-3)(x+3).$$\n\n3. **Evaluate the expression at $x=3$ directly:**\nNumerator: $\sqrt{3+13} - 2\sqrt{3+1} = \sqrt{16} - 2\sqrt{4} = 4 - 2 \times 2 = 4 - 4 = 0.$\nDenominator: $3^2 - 9 = 9 - 9 = 0.$\nSince direct substitution gives $\frac{0}{0}$, we use algebraic manipulation or L'Hôpital's Rule.\n\n4. **Use algebraic manipulation:** Multiply numerator and denominator by the conjugate of the numerator to simplify:\n$$\frac{\sqrt{x+13} - 2\sqrt{x+1}}{(x-3)(x+3)} \times \frac{\sqrt{x+13} + 2\sqrt{x+1}}{\sqrt{x+13} + 2\sqrt{x+1}} = \frac{(x+13) - 4(x+1)}{(x-3)(x+3)(\sqrt{x+13} + 2\sqrt{x+1})}.$$\n\n5. **Simplify the numerator:**\n$$(x+13) - 4(x+1) = x + 13 - 4x - 4 = -3x + 9 = -3(x - 3).$$\n\n6. **Rewrite the expression:**\n$$\frac{-3(x-3)}{(x-3)(x+3)(\sqrt{x+13} + 2\sqrt{x+1})} = \frac{-3}{(x+3)(\sqrt{x+13} + 2\sqrt{x+1})}, \quad x \neq 3.$$\n\n7. **Evaluate the limit as $x \to 3$:**\n$$\lim_{x \to 3} \frac{-3}{(x+3)(\sqrt{x+13} + 2\sqrt{x+1})} = \frac{-3}{(3+3)(\sqrt{3+13} + 2\sqrt{3+1})} = \frac{-3}{6(4 + 2 \times 2)} = \frac{-3}{6(4 + 4)} = \frac{-3}{6 \times 8} = \frac{-3}{48} = -\frac{1}{16}.$$\n\n**Final answer:** $$\boxed{-\frac{1}{16}}.$$