How to find out the values of $\alpha \in \mathbb R$ so that the function $g_\alpha(x)=x^\alpha \ln x$ is uniformly continuous on $(0,\infty)$?
Since for uniform continuity of $g_\alpha(x)$ on $(0,\infty)$, it is necessary that the existence of $\lim_{x \rightarrow 0^+}g_\alpha(x)$, thus we must have $\alpha>0$, but how can we find out an upper bound for $\alpha$?
Also, since for $\alpha>0$, $g_\alpha(x)$ is continuous on $(0,a],~a>0$, we have $g_\alpha(x)$ is uniformly continuous on $(0,a]$, how can we extend the uniformity towards right side?