Find $lim_{x \rightarrow 0^+} x^{\ln x}$

Let $f(x)=x^{\ln x}$. Then
$$\ln f =\ln (x^{\ln x})=\ln x \ln x=(\ln x)^2.$$
So 
$$\lim _{x \rightarrow 0^+}\ln f =\lim _{x \rightarrow 0^+}(\ln x)^2=+\infty.$$
Hence 
$$\ln(\lim_{x \rightarrow 0^+} f(x))=\lim _{x \rightarrow 0^+}\ln f =+\infty.$$ 
Thus
\[\lim_{x \rightarrow 0^+} f(x)=+\infty.\]
So the limit is $\infty$ and not $0$.