Chapter 6 - Inverse Functions - 6.4* General Logarithmic and Exponential Functions - 6.4* Exercise - Page 464: 56

$$0$$

Given $$\lim _{x \rightarrow 0^{+}} x^{-\ln x} $$ Since \begin{aligned} \lim _{x \rightarrow 0^{+}} x^{-\ln x}&= \lim _{x \rightarrow 0^{+}} (e^{-\ln x})^{\ln x}\\ &= \lim _{x \rightarrow 0^{+}} (e^{-\ln x\ln x})\\ &=\lim _{x \rightarrow 0^{+}} (e^{-\ln^2 x})\\ &= e^{-\infty}\\ &=0 \end{aligned}