Answer
$$\frac{2}{5}{\bf{i}} + \frac{1}{3}{\bf{k}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{t \to 2} \left( {\frac{t}{{{t^2} + 1}}\,{\bf{i}} - 4{e^{ - t}}\sin \pi t\,{\bf{j}} + \frac{1}{{\sqrt {4t + 1} }}\,{\bf{k}}} \right) \cr
& {\text{use the sum limits property }}\left( {{\text{see page 805}}} \right) \cr
& = \mathop {\lim }\limits_{t \to 2} \left( {\frac{t}{{{t^2} + 1}}} \right){\bf{i}} - \mathop {\lim }\limits_{t \to 2} \left( {4{e^{ - t}}\sin \pi t} \right){\bf{j}} + \mathop {\lim }\limits_{t \to 2} \left( {\frac{1}{{\sqrt {4t + 1} }}} \right){\bf{k}} \cr
& {\text{evalute each limit substitute }}2{\text{ for }}t \cr
& = \left( {\frac{2}{{{2^2} + 1}}} \right){\bf{i}} - \left( {4{e^{ - 2}}\sin 2\pi } \right){\bf{j}} + \left( {\frac{1}{{\sqrt {4\left( 2 \right) + 1} }}} \right){\bf{k}} \cr
& {\text{simplifying}} \cr
& = \frac{2}{5}{\bf{i}} - \left( 0 \right){\bf{j}} + \left( {\frac{1}{3}} \right){\bf{k}} \cr
& = \frac{2}{5}{\bf{i}} + \frac{1}{3}{\bf{k}} \cr} $$
