1. **State the problem:** Evaluate the limit $$\lim_{x \to 0} \frac{x - 1}{\sqrt{x + 3} - 2}$$. 2. **Recall the formula and approach:** When direct substitution leads to an indeterminate form, we use algebraic manipulation such as rationalizing the denominator. 3. **Check direct substitution:** Substitute $x=0$: $$\frac{0 - 1}{\sqrt{0 + 3} - 2} = \frac{-1}{\sqrt{3} - 2}$$ which is not indeterminate, so the limit is simply this value. 4. **Simplify the denominator:** To express the answer in a simpler form, rationalize the denominator: $$\frac{-1}{\sqrt{3} - 2} \times \frac{\sqrt{3} + 2}{\sqrt{3} + 2} = \frac{-1(\sqrt{3} + 2)}{(\sqrt{3} - 2)(\sqrt{3} + 2)}$$ 5. **Calculate the denominator:** $$(\sqrt{3})^2 - (2)^2 = 3 - 4 = -1$$ 6. **Final expression:** $$\frac{-1(\sqrt{3} + 2)}{-1} = \sqrt{3} + 2$$ 7. **Conclusion:** The limit is $$\boxed{\sqrt{3} + 2}$$. This means as $x$ approaches 0, the function approaches $\sqrt{3} + 2$.