1. **State the problem:** Find the limit as $x$ approaches 0 from the right of the function $f(x) = x^{\sqrt{x}}$. 2. **Recall the limit definition and rewrite the expression:** The function can be rewritten using the exponential and logarithm relationship: $$x^{\sqrt{x}} = e^{\sqrt{x} \ln(x)}$$ 3. **Analyze the exponent:** We need to find the limit of the exponent as $x \to 0^+$: $$\lim_{x \to 0^+} \sqrt{x} \ln(x)$$ 4. **Evaluate the limit of the exponent:** Let $t = \sqrt{x}$, so as $x \to 0^+$, $t \to 0^+$. Then, $$\lim_{t \to 0^+} t \ln(t^2) = \lim_{t \to 0^+} t (2 \ln(t)) = 2 \lim_{t \to 0^+} t \ln(t)$$ 5. **Use substitution and limit properties:** The limit $\lim_{t \to 0^+} t \ln(t) = 0$ because $t$ approaches 0 faster than $\ln(t)$ approaches $-\infty$. 6. **Therefore, the exponent limit is:** $$2 \times 0 = 0$$ 7. **Find the original limit:** $$\lim_{x \to 0^+} x^{\sqrt{x}} = \lim_{x \to 0^+} e^{\sqrt{x} \ln(x)} = e^0 = 1$$ **Final answer:** $$\boxed{1}$$