Solution
= Solution
Write the product as a quotient with indeterminate form $(-\infty)/(+\infty)$. By <L'Hôpital's rule>,
$$
\lim_{x\to0^+}x\ln x=\lim_{x\to0^+}\frac{\ln x}{1/x}=\lim_{x\to0^+}\frac{1/x}{-1/x^2}=\lim_{x\to0^+}(-x)=\boxed{0}.
$$
= Solution
Write the product as a quotient with indeterminate form $(-\infty)/(+\infty)$. By <L'Hôpital's rule>,
$$
\lim_{x\to0^+}x\ln x=\lim_{x\to0^+}\frac{\ln x}{1/x}=\lim_{x\to0^+}\frac{1/x}{-1/x^2}=\lim_{x\to0^+}(-x)=\boxed{0}.
$$