1. **Problem:** Evaluate the limit $$\lim_{x \to 2} \frac{\sqrt{6 - x^2}}{3 - x - 1}$$ 2. **Formula and rules:** To evaluate limits involving square roots and rational expressions, first try direct substitution. If it results in an indeterminate form like $$\frac{0}{0}$$, use algebraic manipulation such as rationalizing the numerator or denominator. 3. **Step 1: Direct substitution** Substitute $$x = 2$$: $$\frac{\sqrt{6 - 2^2}}{3 - 2 - 1} = \frac{\sqrt{6 - 4}}{0} = \frac{\sqrt{2}}{0}$$ The denominator is zero, so direct substitution is undefined. 4. **Step 2: Simplify denominator** Rewrite denominator: $$3 - x - 1 = 2 - x$$ So the limit becomes: $$\lim_{x \to 2} \frac{\sqrt{6 - x^2}}{2 - x}$$ 5. **Step 3: Multiply numerator and denominator by the conjugate of numerator** Multiply by $$\frac{\sqrt{6 - x^2} + \sqrt{2}}{\sqrt{6 - x^2} + \sqrt{2}}$$: $$\lim_{x \to 2} \frac{\sqrt{6 - x^2} (\sqrt{6 - x^2} + \sqrt{2})}{(2 - x)(\sqrt{6 - x^2} + \sqrt{2})} = \lim_{x \to 2} \frac{6 - x^2 - 2}{(2 - x)(\sqrt{6 - x^2} + \sqrt{2})}$$ 6. **Step 4: Simplify numerator** $$6 - x^2 - 2 = 4 - x^2 = (2 - x)(2 + x)$$ 7. **Step 5: Cancel common factor** $$\lim_{x \to 2} \frac{(2 - x)(2 + x)}{(2 - x)(\sqrt{6 - x^2} + \sqrt{2})} = \lim_{x \to 2} \frac{2 + x}{\sqrt{6 - x^2} + \sqrt{2}}$$ 8. **Step 6: Substitute $$x = 2$$** $$\frac{2 + 2}{\sqrt{6 - 4} + \sqrt{2}} = \frac{4}{\sqrt{2} + \sqrt{2}} = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$ **Final answer:** $$\boxed{\sqrt{2}}$$