1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x}$$. 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: $$\frac{\sqrt{x+4} - 2}{x} \times \frac{\sqrt{x+4} + 2}{\sqrt{x+4} + 2} = \frac{(\sqrt{x+4} - 2)(\sqrt{x+4} + 2)}{x(\sqrt{x+4} + 2)}$$ 4. **Simplify numerator using difference of squares:** $$ (\sqrt{x+4})^2 - 2^2 = (x+4) - 4 = x $$ 5. **Rewrite the expression:** $$ \frac{x}{x(\sqrt{x+4} + 2)} = \frac{1}{\sqrt{x+4} + 2} $$ 6. **Evaluate the limit by direct substitution:** $$ \lim_{x \to 0} \frac{1}{\sqrt{x+4} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4} $$ **Final answer:** $$\boxed{\frac{1}{4}}$$