Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to {1^ - }} \frac{{{{\tanh }^{ - 1}}x}}{{\tan \left( {\pi x/2} \right)}} \cr
& {\text{Evaluate the limit by direct substitution}} \cr
& \mathop {\lim }\limits_{x \to {1^ - }} \frac{{{{\tanh }^{ - 1}}x}}{{\tan \left( {\pi x/2} \right)}} = \frac{{{{\tanh }^{ - 1}}\left( 1 \right)}}{{\tan \left( {\pi /2} \right)}} \cr
& {\text{ }} = \frac{\infty }{\infty } \cr
& {\text{Use L'Hopital's Rule}} \cr
& \mathop {\lim }\limits_{x \to {1^ - }} \frac{{{{\tanh }^{ - 1}}x}}{{\tan \left( {\pi x/2} \right)}} = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\frac{d}{{dx}}\left[ {{{\tanh }^{ - 1}}x} \right]}}{{\frac{d}{{dx}}\left[ {\tan \left( {\pi x/2} \right)} \right]}} \cr
& {\text{ }} = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\frac{1}{{1 - {x^2}}}}}{{{{\sec }^2}\left( {\frac{{\pi x}}{2}} \right)\left( {\frac{\pi }{2}} \right)}} \cr
& {\text{ }} = \mathop {\lim }\limits_{x \to {1^ - }} \frac{2}{{\pi \left( {1 - {x^2}} \right){{\sec }^2}\left( {\frac{{\pi x}}{2}} \right)}} \cr
& {\text{Evaluate the limit}} \cr
& {\text{ }} = \mathop {\lim }\limits_{x \to {1^ - }} \frac{2}{{\pi \left( {1 - {1^2}} \right){{\sec }^2}\left( {\frac{\pi }{2}} \right)}} = \frac{2}{\infty } = 0 \cr
& {\text{Therefore}} \cr
& \mathop {\lim }\limits_{x \to {1^ - }} \frac{{{{\tanh }^{ - 1}}x}}{{\tan \left( {\pi x/2} \right)}} = 0 \cr} $$
