1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x}$$. 2. **Recall the formula and rules:** When direct substitution leads to an indeterminate form like $$\frac{0}{0}$$, we use algebraic manipulation such as rationalizing the numerator. 3. **Apply rationalization:** Multiply numerator and denominator by the conjugate of the numerator: $$\frac{\sqrt{x+1} - 1}{x} \times \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1} = \frac{(\sqrt{x+1} - 1)(\sqrt{x+1} + 1)}{x(\sqrt{x+1} + 1)}$$ 4. **Simplify the numerator using difference of squares:** $$= \frac{(x+1) - 1}{x(\sqrt{x+1} + 1)} = \frac{x}{x(\sqrt{x+1} + 1)}$$ 5. **Cancel common factor $x$:** $$= \frac{\cancel{x}}{\cancel{x}(\sqrt{x+1} + 1)} = \frac{1}{\sqrt{x+1} + 1}$$ 6. **Evaluate the limit by direct substitution:** $$\lim_{x \to 0} \frac{1}{\sqrt{x+1} + 1} = \frac{1}{\sqrt{0+1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$$ **Final answer:** $$\boxed{\frac{1}{2}}$$