L'Hôpital's Rule

Let $L = \lim x^x$. Then $\ln L = \lim x \ln x$. Rewrite as $\lim \frac{\ln x}{1/x}$ (form $-\infty/\infty$), apply L'Hopital: $\lim \frac{1/x}{-1/x^2} = \lim (-x) = 0$. So $\ln L = 0 \Rightarrow L = 1$.