1. **State the problem:** Find the limit $$\lim_{x \to 3} \frac{2x - 6}{\sqrt{x+1} - 2}$$. 2. **Identify the indeterminate form:** Substitute $x=3$ directly: $$\frac{2(3) - 6}{\sqrt{3+1} - 2} = \frac{6 - 6}{2 - 2} = \frac{0}{0}$$ which is indeterminate. 3. **Rationalize the denominator:** Multiply numerator and denominator by the conjugate of the denominator: $$\frac{2x - 6}{\sqrt{x+1} - 2} \times \frac{\sqrt{x+1} + 2}{\sqrt{x+1} + 2} = \frac{(2x - 6)(\sqrt{x+1} + 2)}{(\sqrt{x+1} - 2)(\sqrt{x+1} + 2)}$$ 4. **Simplify the denominator using difference of squares:** $$ (\sqrt{x+1})^2 - 2^2 = (x+1) - 4 = x - 3 $$ 5. **Rewrite the expression:** $$ \frac{(2x - 6)(\sqrt{x+1} + 2)}{x - 3} $$ 6. **Factor numerator term:** $$ 2x - 6 = 2(x - 3) $$ 7. **Cancel common factor $(x-3)$:** $$ \frac{2(x - 3)(\sqrt{x+1} + 2)}{x - 3} = 2(\sqrt{x+1} + 2) $$ 8. **Evaluate the limit by substituting $x=3$:** $$ 2(\sqrt{3+1} + 2) = 2(2 + 2) = 2 \times 4 = 8 $$ **Final answer:** 8