Answer
$$\frac{2}{\pi }$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{{\tanh }^{ - 1}}x}}{{\tanh \left( {\pi x/2} \right)}} \cr
& {\text{evaluating the limit}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{{\tanh }^{ - 1}}x}}{{\tan \left( {\pi x/2} \right)}} = \frac{{{{\tanh }^{ - 1}}\left( 0 \right)}}{{\tan \left( 0 \right)}} \cr
& {\text{simplifying}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{{\tanh }^{ - 1}}x}}{{\tan \left( {\pi x/2} \right)}} = \frac{0}{0} \cr
& {\text{using the L'hopital's Rule }} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{{\tanh }^{ - 1}}x}}{{\tan \left( {\pi x/2} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left[ {{{\tanh }^{ - 1}}x} \right]}}{{\frac{d}{{dx}}\left[ {\tan \left( {\pi x/2} \right)} \right]}} \cr
& {\text{solving derivatives}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{1 - {x^2}}}}}{{\frac{\pi }{2}{{\sec }^2}\left( {\frac{{\pi x}}{2}} \right)}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{2}{{\pi \left( {1 - {x^2}} \right){{\sec }^2}\left( {\frac{{\pi x}}{2}} \right)}} \cr
& {\text{evaluate the limit}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{{\pi \left( {1 - {0^2}} \right){{\sec }^2}\left( 0 \right)}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{{\pi \left( 1 \right)\left( 1 \right)}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{\pi } \cr} $$
