1. **State the problem:** Find the limit $$\lim_{x \to +\infty} \frac{\sqrt{x} + \sqrt{x + \sqrt{x}}}{\sqrt{x} + 1}$$ as $x$ approaches positive infinity. 2. **Recall the formula and rules:** When $x \to +\infty$, terms involving $\sqrt{x}$ dominate constants. To simplify limits involving square roots, factor out the highest power of $x$ inside the roots and expressions. 3. **Simplify the numerator:** $$\sqrt{x} + \sqrt{x + \sqrt{x}} = \sqrt{x} + \sqrt{x\left(1 + \frac{\sqrt{x}}{x}\right)} = \sqrt{x} + \sqrt{x} \sqrt{1 + \frac{1}{\sqrt{x}}}$$ 4. **Rewrite the numerator:** $$= \sqrt{x} + \sqrt{x} \sqrt{1 + x^{-1/2}} = \sqrt{x} \left(1 + \sqrt{1 + x^{-1/2}}\right)$$ 5. **Simplify the denominator:** $$\sqrt{x} + 1 = \sqrt{x} \left(1 + \frac{1}{\sqrt{x}}\right)$$ 6. **Rewrite the limit:** $$\lim_{x \to +\infty} \frac{\sqrt{x} \left(1 + \sqrt{1 + x^{-1/2}}\right)}{\sqrt{x} \left(1 + x^{-1/2}\right)} = \lim_{x \to +\infty} \frac{1 + \sqrt{1 + x^{-1/2}}}{1 + x^{-1/2}}$$ 7. **Evaluate the limit as $x \to +\infty$:** Since $x^{-1/2} = \frac{1}{\sqrt{x}} \to 0$, we have $$\sqrt{1 + x^{-1/2}} \to \sqrt{1 + 0} = 1$$ and $$1 + x^{-1/2} \to 1 + 0 = 1$$ 8. **Final limit value:** $$\lim_{x \to +\infty} \frac{1 + 1}{1 + 0} = \frac{2}{1} = 2$$ **Answer:** The limit is $2$.