Answer
$0$
Work Step by Step
Since we have
$$\lim _{x \rightarrow 0+} x^{1 / 2} \ln x=0 . \infty$$
We can apply L’Hôpital’s Rule as follows
$$\lim _{x \rightarrow 0+} x^{1 / 2} \ln x=\lim _{x \rightarrow 0+} \frac{ \ln x}{x^{-1 / 2}}=\lim _{x \rightarrow 0+} \frac{ 1/ x}{(-1/2)x^{-3 / 2}}=-2\lim _{x \rightarrow 0+}x^{1/2}=0.$$
