1. The problem is to find the limit: $$\lim_{x \to 4} \frac{\sqrt{1+2x} - 3}{\sqrt{x} - 2}$$ 2. This is a limit of the form $\frac{0}{0}$, so we can use algebraic manipulation to simplify it. 3. Multiply numerator and denominator by the conjugate of the numerator: $$\frac{\sqrt{1+2x} - 3}{\sqrt{x} - 2} \cdot \frac{\sqrt{1+2x} + 3}{\sqrt{1+2x} + 3} = \frac{(1+2x) - 9}{(\sqrt{x} - 2)(\sqrt{1+2x} + 3)}$$ 4. Simplify the numerator: $$1 + 2x - 9 = 2x - 8 = 2(x - 4)$$ 5. Rewrite the limit expression: $$\lim_{x \to 4} \frac{2(x - 4)}{(\sqrt{x} - 2)(\sqrt{1+2x} + 3)}$$ 6. Notice that $\sqrt{x} - 2$ in the denominator can be factored with $x - 4$ in the numerator by using the difference of squares: $$x - 4 = (\sqrt{x} - 2)(\sqrt{x} + 2)$$ 7. Substitute this into the numerator: $$\lim_{x \to 4} \frac{2(\sqrt{x} - 2)(\sqrt{x} + 2)}{(\sqrt{x} - 2)(\sqrt{1+2x} + 3)}$$ 8. Cancel $\sqrt{x} - 2$ terms: $$\lim_{x \to 4} \frac{2(\sqrt{x} + 2)}{\sqrt{1+2x} + 3}$$ 9. Now substitute $x = 4$: $$\frac{2(\sqrt{4} + 2)}{\sqrt{1 + 2 \cdot 4} + 3} = \frac{2(2 + 2)}{\sqrt{9} + 3} = \frac{2 \cdot 4}{3 + 3} = \frac{8}{6} = \frac{4}{3}$$ Final answer: $$\boxed{\frac{4}{3}}$$