1. **State the problem:** We want to find the limit as $x$ approaches 1 of the expression $$\frac{\sqrt{x+3} - 2}{3 - \sqrt{x+8}}$$ 2. **Evaluate the direct substitution:** Substitute $x=1$: Numerator: $\sqrt{1+3} - 2 = \sqrt{4} - 2 = 2 - 2 = 0$ Denominator: $3 - \sqrt{1+8} = 3 - \sqrt{9} = 3 - 3 = 0$ We get an indeterminate form $\frac{0}{0}$. 3. **Rationalize numerator and denominator separately:** Multiply numerator and denominator by the conjugates to simplify. Multiply numerator and denominator by $\frac{\sqrt{x+3}+2}{\sqrt{x+3}+2}$: $$\frac{\sqrt{x+3} - 2}{3 - \sqrt{x+8}} \cdot \frac{\sqrt{x+3} + 2}{\sqrt{x+3} + 2} = \frac{(x+3) - 4}{(3 - \sqrt{x+8})(\sqrt{x+3} + 2)} = \frac{x - 1}{(3 - \sqrt{x+8})(\sqrt{x+3} + 2)}$$ 4. **Next, rationalize the denominator factor $3 - \sqrt{x+8}$ by multiplying numerator and denominator by $3 + \sqrt{x+8}$:** $$\frac{x - 1}{(3 - \sqrt{x+8})(\sqrt{x+3} + 2)} \cdot \frac{3 + \sqrt{x+8}}{3 + \sqrt{x+8}} = \frac{(x - 1)(3 + \sqrt{x+8})}{(9 - (x + 8))(\sqrt{x+3} + 2)} = \frac{(x - 1)(3 + \sqrt{x+8})}{(1 - x)(\sqrt{x+3} + 2)}$$ 5. **Simplify numerator and denominator:** Note $9-(x+8) = 1 - x$. Rewrite denominator $1 - x = -(x - 1)$: $$\frac{(x - 1)(3 + \sqrt{x+8})}{-(x - 1)(\sqrt{x+3} + 2)} = \frac{3 + \sqrt{x+8}}{-(\sqrt{x+3} + 2)}$$ 6. **Cancel $(x - 1)$ terms and evaluate limit $x \to 1$:** $$\lim_{x \to 1} \frac{3 + \sqrt{x+8}}{-(\sqrt{x+3} + 2)} = \frac{3 + \sqrt{1+8}}{-(\sqrt{1+3} + 2)} = \frac{3 + 3}{-(2 + 2)} = \frac{6}{-4} = -\frac{3}{2}$$ **Final answer:** $$\boxed{-\frac{3}{2}}$$