1) \( \lim _{x \rightarrow 0^{+}} x \ln x \)

To solve the limit \( \lim _{x \rightarrow 0^{+}} x \ln x \), we can rewrite it in a form that's easier to analyze. As \( x \) approaches 0 from the right, \( \ln x \) approaches \( -\infty \). Therefore, we can express the limit as: \[ \lim_{x \to 0^{+}} x \ln x = \lim_{x \to 0^{+}} \frac{\ln x}{\frac{1}{x}}. \] Here, both the numerator and denominator trend towards negative infinity, thereby allowing us to use L'Hôpital's Rule: \[ \text{Differentiate the numerator: } \frac{d}{dx}(\ln x) = \frac{1}{x}, \quad \text{and differentiate the denominator: } \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}. \] Applying L'Hôpital's Rule gives us: \[ \lim_{x \to 0^{+}} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^{+}} -x = 0. \] Thus, the limit is: \[ \lim _{x \rightarrow 0^{+}} x \ln x = 0. \] So the final answer is \( 0 \).