1. **Problem statement:** Find the limit as $x \to 0$ of $$\sqrt{3x + 5} - \sqrt{3x - 3}.$$\n\n2. **Formula and approach:** When dealing with limits involving differences of square roots, a common technique is to multiply and divide by the conjugate to simplify the expression. The conjugate of $$\sqrt{3x + 5} - \sqrt{3x - 3}$$ is $$\sqrt{3x + 5} + \sqrt{3x - 3}.$$\n\n3. **Multiply numerator and denominator by the conjugate:**\n$$\lim_{x \to 0} \frac{\sqrt{3x + 5} - \sqrt{3x - 3}}{1} \times \frac{\sqrt{3x + 5} + \sqrt{3x - 3}}{\sqrt{3x + 5} + \sqrt{3x - 3}} = \lim_{x \to 0} \frac{(\sqrt{3x + 5})^2 - (\sqrt{3x - 3})^2}{\sqrt{3x + 5} + \sqrt{3x - 3}}.$$\n\n4. **Simplify the numerator:**\n$$= \lim_{x \to 0} \frac{3x + 5 - (3x - 3)}{\sqrt{3x + 5} + \sqrt{3x - 3}} = \lim_{x \to 0} \frac{3x + 5 - 3x + 3}{\sqrt{3x + 5} + \sqrt{3x - 3}} = \lim_{x \to 0} \frac{8}{\sqrt{3x + 5} + \sqrt{3x - 3}}.$$\n\n5. **Evaluate the limit by substituting $x=0$ in the denominator:**\n$$= \frac{8}{\sqrt{5} + \sqrt{-3}}.$$\n\n6. **Note:** Since $\sqrt{-3}$ is not a real number, the limit does not exist in the real numbers.\n\n**Final answer:** The limit does not exist in the real number system because the expression involves the square root of a negative number as $x \to 0$.