Answer
$$2{\bf{i}} + {\bf{j}} + {\bf{k}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{t \to 0} \left( {\frac{{\sin 2t}}{t} + {e^{ - t}}{\bf{j}} + {e^t}{\bf{k}}} \right) \cr
& {\text{Evaluate the limit}} \cr
& {\text{As t approaches }}0,{\text{ the limit is}} \cr
& {\text{ = }}\left[ {\mathop {\lim }\limits_{t \to 0} \frac{{\sin 2t}}{t}} \right]{\bf{i}} + \left[ {\mathop {\lim }\limits_{t \to 0} {e^{ - t}}} \right]{\bf{j}} + \left[ {\mathop {\lim }\limits_{t \to 0} {e^t}} \right]{\bf{k}} \cr
& {\text{ = }}2\left[ {\mathop {\lim }\limits_{t \to 0} \frac{{\sin 2t}}{{2t}}} \right]{\bf{i}} + \left[ {{e^{ - 0}}} \right]{\bf{j}} + \left[ {{e^0}} \right]{\bf{k}} \cr
& = 2\left( 1 \right){\bf{i}} + \left( 1 \right){\bf{j}} + \left( 1 \right){\bf{k}} \cr
& = 2{\bf{i}} + {\bf{j}} + {\bf{k}} \cr} $$
