Answer
$$1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } {\left( {\ln x} \right)^{2/x}} \cr
& {\text{Evaluating the limit}} \cr
& \mathop {\lim }\limits_{x \to \infty } {\left( {\ln x} \right)^{2/x}} = {\left( {\ln \infty } \right)^{2/\infty }} = {\infty ^0} \cr
& {\text{This limit has the form }}{\infty ^0}{\text{ }}\left( {{\text{See example 6, page 562}}} \right) \cr
& {\text{Let }}y = \mathop {\lim }\limits_{x \to \infty } {\left( {\ln x} \right)^{2/x}} \cr
& {\text{Take the natural log of each side}} \cr
& \ln y = \ln \left[ {\mathop {\lim }\limits_{x \to \infty } {{\left( {\ln x} \right)}^{2/x}}} \right] \cr
& {\text{Continuity}} \cr
& \ln y = \left[ {\mathop {\lim }\limits_{x \to \infty } \ln {{\left( {\ln x} \right)}^{2/x}}} \right] \cr
& {\text{Power property of logarithms}} \cr
& \ln y = \mathop {\lim }\limits_{x \to \infty } \frac{{2\ln \left( {\ln x} \right)}}{x} \cr
& {\text{Evaluating }} \cr
& \ln y = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{2\ln \left( {\ln \infty } \right)}}{\infty }} \right] = \frac{\infty }{\infty } \cr
& {\text{Using L'Hopital's rule}} \cr
& \ln y = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{\frac{d}{{dx}}\left[ {2\ln \left( {\ln x} \right)} \right]}}{{\frac{d}{{dx}}\left[ x \right]}}} \right] \cr
& \ln y = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{2}{{x\ln x}}} \right] \cr
& {\text{Evaluating the limit}} \cr
& \ln y = \frac{2}{\infty } \cr
& \ln y = 0 \cr
& y = 1 \cr
& {\text{Therefore}} \cr
& y = \mathop {\lim }\limits_{x \to \infty } {\left( {\ln x} \right)^{2/x}} \cr
& \mathop {\lim }\limits_{x \to \infty } {\left( {\ln x} \right)^{2/x}} = 1 \cr} $$
