So, we're faced with this:

$\displaystyle \lim_{x \rightarrow 0^+} \, x\ln (x) \, .$

Since we get an indeterminate form when attempting to evaluate at $\displaystyle x = 0$, we resort to L'Hôspitals rule. The derivative of $\displaystyle x\ln (x)$ is $\displaystyle \ln (x) + 1$, therefore

$\displaystyle \lim_{x \rightarrow 0^+} \, x\ln (x) = \lim_{x \rightarrow 0^+} \, \ln (x) + 1 \, .$

I am stuck here, as it eludes me how this approaches 0? I get negative infinity, but apparently that's incorrect.