1. **State the problem:** Find the limit $$\lim_{x \to 0} \left( \sqrt{3x + 5} - \sqrt{3x - 3} \right).$$ 2. **Recall the formula and technique:** When dealing with limits involving differences of square roots, multiply and divide by the conjugate to simplify. 3. **Multiply by the conjugate:** $$\lim_{x \to 0} \frac{\left( \sqrt{3x + 5} - \sqrt{3x - 3} \right) \left( \sqrt{3x + 5} + \sqrt{3x - 3} \right)}{\sqrt{3x + 5} + \sqrt{3x - 3}} = \lim_{x \to 0} \frac{(3x + 5) - (3x - 3)}{\sqrt{3x + 5} + \sqrt{3x - 3}}.$$ 4. **Simplify the numerator:** $$(3x + 5) - (3x - 3) = 3x + 5 - 3x + 3 = 8.$$ 5. **Rewrite the limit:** $$\lim_{x \to 0} \frac{8}{\sqrt{3x + 5} + \sqrt{3x - 3}}.$$ 6. **Evaluate the denominator at $x=0$:** $$\sqrt{3(0) + 5} + \sqrt{3(0) - 3} = \sqrt{5} + \sqrt{-3}.$$ 7. **Note:** $\sqrt{-3}$ is not a real number, so the limit does not exist in the real numbers. **Final answer:** The limit does not exist in the real number system because the expression involves the square root of a negative number as $x$ approaches 0.