1. **State the problem:** We want to find the limit of the function $$f(x) = \sqrt{x} - 1 - \sqrt{x} \ln(x)$$ as $$x$$ approaches $$0^+$$ (from the right side). 2. **Recall the behavior of components near 0:** - $$\sqrt{x}$$ approaches 0 as $$x \to 0^+$$. - $$\ln(x)$$ approaches $$-\infty$$ as $$x \to 0^+$$. 3. **Rewrite the function to analyze the limit:** $$f(x) = \sqrt{x} - 1 - \sqrt{x} \ln(x) = (\sqrt{x} - 1) - \sqrt{x} \ln(x)$$ 4. **Analyze each term separately:** - The term $$\sqrt{x} - 1$$ approaches $$-1$$ as $$x \to 0^+$$ because $$\sqrt{x} \to 0$$. - The term $$-\sqrt{x} \ln(x)$$ is an indeterminate form of type $$0 \cdot \infty$$. 5. **Evaluate the limit of $$-\sqrt{x} \ln(x)$$:** Set $$t = \sqrt{x}$$, so as $$x \to 0^+$$, $$t \to 0^+$$. Then, $$-\sqrt{x} \ln(x) = -t \ln(t^2) = -t (2 \ln t) = -2t \ln t$$ 6. **Use the known limit:** $$\lim_{t \to 0^+} t \ln t = 0$$. Therefore, $$\lim_{t \to 0^+} -2t \ln t = 0$$. 7. **Combine the limits:** $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (\sqrt{x} - 1) - \lim_{x \to 0^+} \sqrt{x} \ln(x) = -1 + 0 = -1$$. **Final answer:** $$\boxed{-1}$$