1. **Problem:** Find the limit $$\lim_{x \to 0} \frac{\sqrt{x + a} - \sqrt{a}}{x}$$ where $a > 0$. 2. **Formula and rule:** To evaluate limits involving square roots, multiply numerator and denominator by the conjugate to simplify. 3. **Work:** Multiply numerator and denominator by $$\sqrt{x + a} + \sqrt{a}$$: $$\lim_{x \to 0} \frac{\sqrt{x + a} - \sqrt{a}}{x} \times \frac{\sqrt{x + a} + \sqrt{a}}{\sqrt{x + a} + \sqrt{a}} = \lim_{x \to 0} \frac{(x + a) - a}{x(\sqrt{x + a} + \sqrt{a})} = \lim_{x \to 0} \frac{x}{x(\sqrt{x + a} + \sqrt{a})}$$ 4. Simplify numerator and denominator: $$= \lim_{x \to 0} \frac{1}{\sqrt{x + a} + \sqrt{a}}$$ 5. Substitute $x = 0$: $$= \frac{1}{\sqrt{a} + \sqrt{a}} = \frac{1}{2\sqrt{a}}$$ **Final answer:** $$\boxed{\frac{1}{2\sqrt{a}}}$$