1. We are asked to find the limit: $$\lim_{x \to 4} \frac{\sqrt{4+x} - \sqrt{2x}}{x-4}$$ 2. This is an indeterminate form of type $\frac{0}{0}$ because substituting $x=4$ gives $\frac{\sqrt{8} - \sqrt{8}}{0} = \frac{0}{0}$. 3. To resolve this, we use the conjugate to rationalize the numerator. Multiply numerator and denominator by $$\sqrt{4+x} + \sqrt{2x}$$: $$\frac{\sqrt{4+x} - \sqrt{2x}}{x-4} \times \frac{\sqrt{4+x} + \sqrt{2x}}{\sqrt{4+x} + \sqrt{2x}} = \frac{(4+x) - 2x}{(x-4)(\sqrt{4+x} + \sqrt{2x})}$$ 4. Simplify the numerator: $$4 + x - 2x = 4 - x$$ 5. So the expression becomes: $$\frac{4 - x}{(x-4)(\sqrt{4+x} + \sqrt{2x})}$$ 6. Notice that $4 - x = -(x - 4)$, so: $$\frac{-(x - 4)}{(x-4)(\sqrt{4+x} + \sqrt{2x})} = \frac{-1}{\sqrt{4+x} + \sqrt{2x}}$$ 7. Now, take the limit as $x \to 4$: $$\lim_{x \to 4} \frac{-1}{\sqrt{4+4} + \sqrt{2 \times 4}} = \frac{-1}{\sqrt{8} + \sqrt{8}} = \frac{-1}{2\sqrt{8}}$$ 8. Simplify $\sqrt{8} = 2\sqrt{2}$: $$\frac{-1}{2 \times 2 \sqrt{2}} = \frac{-1}{4 \sqrt{2}}$$ 9. Final answer: $$\boxed{\frac{-1}{4 \sqrt{2}}}$$