Answer
$$1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to {0^ + }} {\left| {\ln x} \right|^x} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {\left| {\ln x} \right|^x} = \mathop {\lim }\limits_{x \to {0^ + }} {\left| {\ln 0} \right|^x} = {\infty ^0} \cr
& {\text{This limit has the form }}{\infty ^0}.{\text{ }}\left( {{\text{Using theorem 2}}{\text{.12}}} \right) \cr
& {\left| {\ln x} \right|^x} = {e^{x\ln \left( {\ln x} \right)}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {\left| {\ln x} \right|^x} = \mathop {\lim }\limits_{x \to {0^ + }} {e^{x\ln \left( {\ln x} \right)}} = {e^{\mathop {\lim }\limits_{x \to {0^ + }} x\ln \left( {\ln x} \right)}} \cr
& {\text{The first step is to evaluate }} \cr
& L = \mathop {\lim }\limits_{x \to {0^ + }} x\ln \left( {\ln x} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln \left( {\ln x} \right)}}{{\frac{1}{x}}} = \frac{\infty }{\infty } \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln \left( {\ln x} \right)}}{{\frac{1}{x}}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{1}{{x\ln x}}}}{{ - \frac{1}{{{x^2}}}}} = - \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{x^2}}}{{x\ln x}} = - \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{{\ln x}} \cr
& - \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{{\ln x}} = - \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{{\ln x}} = \frac{0}{\infty } = 0 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {\left| {\ln x} \right|^x} = \mathop {\lim }\limits_{x \to {0^ + }} {e^{x\ln \left( {\ln x} \right)}} = {e^{\mathop {\lim }\limits_{x \to {0^ + }} x\ln \left( {\ln x} \right)}} = {e^0} = 1 \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {\left| {\ln x} \right|^x} = 1 \cr} $$
