1. **State the problem:** We want to find the limit as $x$ approaches positive infinity of the expression $$\sqrt{x^2 + \sqrt{x^2 + 1}} - x.$$\n\n2. **Recall the formula and approach:** When dealing with limits involving square roots and large $x$, a common technique is to rationalize the expression to simplify it.\n\n3. **Rationalize the expression:** Multiply and divide by the conjugate $$\sqrt{x^2 + \sqrt{x^2 + 1}} + x$$ to get\n$$\left(\sqrt{x^2 + \sqrt{x^2 + 1}} - x\right) \cdot \frac{\sqrt{x^2 + \sqrt{x^2 + 1}} + x}{\sqrt{x^2 + \sqrt{x^2 + 1}} + x} = \frac{\left(x^2 + \sqrt{x^2 + 1}\right) - x^2}{\sqrt{x^2 + \sqrt{x^2 + 1}} + x} = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2 + \sqrt{x^2 + 1}} + x}.$$\n\n4. **Simplify numerator:** For large $x$, $$\sqrt{x^2 + 1} = \sqrt{x^2\left(1 + \frac{1}{x^2}\right)} = x\sqrt{1 + \frac{1}{x^2}}.$$\nUsing binomial expansion for large $x$, $$\sqrt{1 + \frac{1}{x^2}} \approx 1 + \frac{1}{2x^2}.$$\nSo, $$\sqrt{x^2 + 1} \approx x + \frac{1}{2x}.$$\n\n5. **Simplify denominator:** Similarly, inside the denominator, $$\sqrt{x^2 + \sqrt{x^2 + 1}} = \sqrt{x^2 + x + \frac{1}{2x}} \approx \sqrt{x^2 + x}$$ for large $x$.\nFactor $x^2$:\n$$\sqrt{x^2 + x} = x\sqrt{1 + \frac{1}{x}}.$$\nUsing binomial expansion again, $$\sqrt{1 + \frac{1}{x}} \approx 1 + \frac{1}{2x}.$$\nSo, $$\sqrt{x^2 + \sqrt{x^2 + 1}} \approx x + \frac{1}{2}.$$\n\n6. **Substitute approximations back:**\n$$\frac{\sqrt{x^2 + 1}}{\sqrt{x^2 + \sqrt{x^2 + 1}} + x} \approx \frac{x + \frac{1}{2x}}{\left(x + \frac{1}{2}\right) + x} = \frac{x + \frac{1}{2x}}{2x + \frac{1}{2}}.$$\n\n7. **Divide numerator and denominator by $x$ to simplify:**\n$$\frac{\cancel{x} + \frac{1}{2\cancel{x}}}{2\cancel{x} + \frac{1}{2}} = \frac{1 + \frac{1}{2x^2}}{2 + \frac{1}{2x}}.$$\n\n8. **Take the limit as $x \to +\infty$:**\nAs $x$ grows large, terms with $\frac{1}{x}$ or $\frac{1}{x^2}$ approach zero, so\n$$\lim_{x \to +\infty} \frac{1 + \frac{1}{2x^2}}{2 + \frac{1}{2x}} = \frac{1 + 0}{2 + 0} = \frac{1}{2}.$$\n\n**Final answer:**\n$$\boxed{\frac{1}{2}}.$$