x ln x at zero, and x^x

1. The product $x\ln x$ has the form $0\times(-\infty)$, which is indeterminate. Put $y=\frac1x$; when $x\to0^+$, $y\to+\infty$, and
$$x\ln x=\frac1y\ln\frac1y=-\frac{\ln y}{y}\longrightarrow0.$$

2. For $x>0$, $x^x=e^{x\ln x}$. The exponent tends to $0$, and $\exp$ is continuous, so
$$\lim_{x\to0^+}x^x=e^0=1.$$

Remark. This is why one defines $0^0=1$ in many contexts: the function $x^x$ extends continuously to $0$ with the value $1$.