1. **State the problem:** Calculate the limit $$\lim_{x \to 1} \frac{\sqrt{3x - 1} - \sqrt{3 - x}}{x - 1}$$ 2. **Recall the formula and technique:** When a limit results in an indeterminate form like $\frac{0}{0}$, we can use algebraic manipulation such as multiplying by the conjugate to simplify. 3. **Check direct substitution:** Substitute $x=1$: $$\sqrt{3(1) - 1} - \sqrt{3 - 1} = \sqrt{2} - \sqrt{2} = 0$$ and denominator $1 - 1 = 0$, so the limit is of the form $\frac{0}{0}$. 4. **Multiply numerator and denominator by the conjugate:** $$\frac{\sqrt{3x - 1} - \sqrt{3 - x}}{x - 1} \times \frac{\sqrt{3x - 1} + \sqrt{3 - x}}{\sqrt{3x - 1} + \sqrt{3 - x}} = \frac{(3x - 1) - (3 - x)}{(x - 1)(\sqrt{3x - 1} + \sqrt{3 - x})}$$ 5. **Simplify the numerator:** $$(3x - 1) - (3 - x) = 3x - 1 - 3 + x = 4x - 4 = 4(x - 1)$$ 6. **Cancel the common factor $(x - 1)$:** $$\frac{4(x - 1)}{(x - 1)(\sqrt{3x - 1} + \sqrt{3 - x})} = \frac{4}{\sqrt{3x - 1} + \sqrt{3 - x}}$$ 7. **Evaluate the limit by substituting $x=1$ now:** $$\frac{4}{\sqrt{3(1) - 1} + \sqrt{3 - 1}} = \frac{4}{\sqrt{2} + \sqrt{2}} = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$ **Final answer:** $$\lim_{x \to 1} \frac{\sqrt{3x - 1} - \sqrt{3 - x}}{x - 1} = \sqrt{2}$$