Answer
$$\frac{1}{{{e^{2/\pi }}}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{2}{\pi }{{\tan }^{ - 1}}x} \right)^x} \cr
& {\text{Evaluate }} \cr
& \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{2}{\pi }{{\tan }^{ - 1}}x} \right)^x} = \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{2}{\pi }{{\tan }^{ - 1}}\left( \infty \right)} \right)^\infty } = {\left( {\frac{2}{\pi }\left( {\frac{\pi }{2}} \right)} \right)^\infty } = {1^\infty } \cr
& {\text{This limit has the form }}{1^\infty }.{\text{ }}\left( {{\text{Using theorem 2}}{\text{.12}}} \right) \cr
& {\text{Noting that }}{\left( {\frac{2}{\pi }{{\tan }^{ - 1}}x} \right)^x} = {e^{x\ln \left( {\frac{2}{\pi }{{\tan }^{ - 1}}x} \right)}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{2}{\pi }{{\tan }^{ - 1}}x} \right)^x} = \mathop {\lim }\limits_{x \to {0^ + }} {e^{x\ln \left( {\frac{2}{\pi }{{\tan }^{ - 1}}x} \right)}} = {e^{\mathop {\lim }\limits_{x \to \infty } x\ln \left( {\frac{2}{\pi }{{\tan }^{ - 1}}x} \right)}} \cr
& {\text{The first step is to evaluate }} \cr
& L = \mathop {\lim }\limits_{x \to \infty } x\ln \left( {\frac{2}{\pi }{{\tan }^{ - 1}}x} \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{\ln \left( {\frac{2}{\pi }{{\tan }^{ - 1}}x} \right)}}{{\frac{1}{x}}} = \frac{0}{0} \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {\ln \left( {\frac{2}{\pi }{{\tan }^{ - 1}}x} \right)} \right]}}{{\frac{d}{{dx}}\left[ {\frac{1}{x}} \right]}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{\frac{2}{\pi }\left( {\frac{1}{{1 + {x^2}}}} \right)}}{{\frac{2}{\pi }{{\tan }^{ - 1}}x}}}}{{ - \frac{1}{{{x^2}}}}} \cr
& L = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{1}{{\left( {1 + {x^2}} \right){{\tan }^{ - 1}}x}}}}{{ - \frac{1}{{{x^2}}}}} \cr
& L = - \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}}}{{\left( {1 + {x^2}} \right){{\tan }^{ - 1}}x}} \cr
& L = - \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{{x^2}}}{{{x^2}}}}}{{\frac{{{{\tan }^{ - 1}}x}}{{{x^2}}} + \frac{{{x^2}{{\tan }^{ - 1}}x}}{{{x^2}}}}} \cr
& L = - \mathop {\lim }\limits_{x \to \infty } \frac{1}{{\frac{{{{\tan }^{ - 1}}x}}{{{x^2}}} + {{\tan }^{ - 1}}x}} \cr
& {\text{Evaluating}} \cr
& L = - \frac{1}{{\frac{{{{\tan }^{ - 1}}\infty }}{{{\infty ^2}}} + {{\tan }^{ - 1}}\left( \infty \right)}} = - \frac{1}{{0 + \pi /2}} = - \frac{2}{\pi } \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{2}{\pi }{{\tan }^{ - 1}}x} \right)^x} = \mathop {\lim }\limits_{x \to {0^ + }} {e^{x\ln \left( {\frac{2}{\pi }{{\tan }^{ - 1}}x} \right)}} = {e^{\mathop {\lim }\limits_{x \to \infty } x\ln \left( {\frac{2}{\pi }{{\tan }^{ - 1}}x} \right)}} = {e^{ - \frac{2}{\pi }}} \cr
& \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{2}{\pi }{{\tan }^{ - 1}}x} \right)^x} = \frac{1}{{{e^{2/\pi }}}} \approx 0.53 \cr} $$
