6.58. $\lim _{x \rightarrow+0} x^{x}$.
Solution.

$\lim _{x \rightarrow+0} x^{x}=e^{\lim _{x \rightarrow 0} x \ln x}=e^{\lim _{x \rightarrow 0 \rightarrow 0(1)}\left(\frac{(\ln x)'}{(1 / x)'}\right)}=e^{\lim _{x \rightarrow-0} \frac{1 / x}{-1 / x^{2}}}=e^{\lim _{x \rightarrow+0}(-x)}=e^{0}=1$.