1. **State the problem:** We want to find the limit $$\lim_{x \to 0^+} x \ln x$$. 2. **Recall the behavior:** As $x$ approaches $0$ from the right, $\ln x$ approaches $-\infty$, and $x$ approaches $0$. This is an indeterminate form of type $0 \cdot (-\infty)$. 3. **Rewrite the expression:** To handle this indeterminate form, rewrite the product as a quotient: $$x \ln x = \frac{\ln x}{\frac{1}{x}}$$ 4. **Apply L'Hôpital's Rule:** Since as $x \to 0^+$, $\ln x \to -\infty$ and $\frac{1}{x} \to \infty$, the quotient is of the form $\frac{-\infty}{\infty}$, suitable for L'Hôpital's Rule. 5. **Differentiate numerator and denominator:** $$\frac{d}{dx}(\ln x) = \frac{1}{x}$$ $$\frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}$$ 6. **Form the new limit:** $$\lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} \frac{1}{x} \cdot \frac{-x^2}{1} = \lim_{x \to 0^+} -x = 0$$ 7. **Conclusion:** The limit is $$\boxed{0}$$. This means that although $\ln x$ goes to negative infinity, the factor $x$ goes to zero fast enough to make the product approach zero.