Answer
$$ - 2{\bf{j}} + \frac{\pi }{2}{\bf{k}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{t \to \infty } \left( {{e^{ - t}}\,{\bf{i}} - \frac{{2t}}{{t + 1}}\,{\bf{j}} + {{\tan }^{ - 1}}t\,{\bf{k}}} \right) \cr
& {\text{use the sum limits property }}\left( {{\text{see page 805}}} \right) \cr
& = \mathop {\lim }\limits_{t \to \infty } \left( {{e^{ - t}}\,} \right){\bf{i}} - \mathop {\lim }\limits_{t \to \infty } \left( {\frac{{2t}}{{t + 1}}} \right){\bf{j}} + \mathop {\lim }\limits_{t \to \infty } \left( {{{\tan }^{ - 1}}t} \right){\bf{k}} \cr
& {\text{write }}\frac{{2t}}{{t + 1}}{\text{ as 2}} - \frac{2}{{t + 1}} \cr
& = \mathop {\lim }\limits_{t \to \infty } \left( {{e^{ - t}}\,} \right){\bf{i}} - \mathop {\lim }\limits_{t \to \infty } \left( {{\text{2}} - \frac{2}{{t + 1}}} \right){\bf{j}} + \mathop {\lim }\limits_{t \to \infty } \left( {{{\tan }^{ - 1}}t} \right){\bf{k}} \cr
& {\text{evalute each limit substitute }}\infty {\text{ for }}t \cr
& = \left( {{e^{ - \infty }}\,} \right){\bf{i}} - \left( {{\text{2}} - \frac{2}{{\infty + 1}}} \right){\bf{j}} + \left( {{{\tan }^{ - 1}}\infty } \right){\bf{k}} \cr
& {\text{simplifying}} \cr
& = \left( 0 \right){\bf{i}} - \left( {\text{2}} \right){\bf{j}} + \left( {\frac{\pi }{2}} \right){\bf{k}} \cr
& = - 2{\bf{j}} + \frac{\pi }{2}{\bf{k}} \cr} $$
