1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x}$$. 2. **Recall the formula and rules:** When a limit results in an indeterminate form like $$\frac{0}{0}$$, we can use algebraic manipulation such as rationalizing the numerator. 3. **Check the direct substitution:** Substitute $x=0$: $$\frac{\sqrt{0+4} - 2}{0} = \frac{2 - 2}{0} = \frac{0}{0}$$ which is indeterminate. 4. **Rationalize the numerator:** Multiply numerator and denominator by the conjugate of the numerator: $$\frac{\sqrt{x+4} - 2}{x} \cdot \frac{\sqrt{x+4} + 2}{\sqrt{x+4} + 2} = \frac{(\sqrt{x+4} - 2)(\sqrt{x+4} + 2)}{x(\sqrt{x+4} + 2)}$$ 5. **Simplify the numerator using difference of squares:** $$ (\sqrt{x+4})^2 - 2^2 = (x+4) - 4 = x $$ 6. **Rewrite the expression:** $$ \frac{x}{x(\sqrt{x+4} + 2)} $$ 7. **Cancel common factor $x$:** $$ \frac{\cancel{x}}{\cancel{x}(\sqrt{x+4} + 2)} = \frac{1}{\sqrt{x+4} + 2} $$ 8. **Evaluate the limit by substituting $x=0$:** $$ \frac{1}{\sqrt{0+4} + 2} = \frac{1}{2 + 2} = \frac{1}{4} $$ **Final answer:** $$ \lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x} = \frac{1}{4} $$