Answer
$$2$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } x{\tan ^{ - 1}}\left( {\frac{2}{x}} \right) \cr
& {\text{evaluating the limit, we get:}} \cr
& \mathop {\lim }\limits_{x \to \infty } x{\tan ^{ - 1}}\left( {\frac{2}{x}} \right) = \left( \infty \right){\tan ^{ - 1}}\left( {\frac{1}{\infty }} \right) \cr
& \mathop {\lim }\limits_{x \to \infty } x{\tan ^{ - 1}}\left( {\frac{2}{x}} \right) = \infty \cdot 0 \cr
& {\text{converted to the form }}\frac{0}{0}.{\text{ Let }}h = \frac{1}{x} \cr
& \mathop {\lim }\limits_{x \to \infty } x{\tan ^{ - 1}}\left( {\frac{2}{x}} \right) = \mathop {\lim }\limits_{h \to {0^ + }} \left( {\frac{1}{h}} \right){\tan ^{ - 1}}\left( {2h} \right) \cr
& = \mathop {\lim }\limits_{h \to {0^ + }} \left( {\frac{{{{\tan }^{ - 1}}\left( {2h} \right)}}{h}} \right) \cr
& {\text{evaluating the limit, we get:}} \cr
& = \left( {\frac{{{{\tan }^{ - 1}}\left( 0 \right)}}{0}} \right) = \frac{0}{0} \cr
& {\text{The limit is }}\frac{0}{0}{\text{, so}}{\text{ we can apply l'Hopital's Rule}} \cr
& = \mathop {\lim }\limits_{h \to {0^ + }} \left( {\frac{{d/dh\left( {{{\tan }^{ - 1}}\left( {2h} \right)} \right)}}{{d/dh\left( h \right)}}} \right) \cr
& = \mathop {\lim }\limits_{h \to {0^ + }} \left( {\frac{{\frac{2}{{1 + {{\left( {2h} \right)}^2}}}}}{1}} \right) \cr
& {\text{evaluating the limit, we get:}} \cr
& = \frac{2}{{1 + {{\left( {2\left( 0 \right)} \right)}^2}}} = 2 \cr
& {\text{Then}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to \infty } x{\tan ^{ - 1}}\left( {\frac{2}{x}} \right) = 2 \cr} $$
