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