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