Answer
$$4{\bf{i}} + 3{\bf{j}} - {\bf{k}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{t \to \ln 2} \left( {2{e^t}\,{\bf{i}} + 6{e^{ - t}}\,{\bf{j}} - 4{e^{ - 2t}}\,{\bf{k}}} \right) \cr
& {\text{use the sum limits property }}\left( {{\text{see page 805}}} \right) \cr
& = \mathop {\lim }\limits_{t \to \ln 2} \left( {2{e^t}} \right){\bf{i}} + \mathop {\lim }\limits_{t \to \ln 2} \left( {6{e^{ - t}}} \right){\bf{j}} - \mathop {\lim }\limits_{t \to \ln 2} \left( {4{e^{ - 2t}}} \right){\bf{k}} \cr
& = 2\mathop {\lim }\limits_{t \to \ln 2} \left( {{e^t}} \right){\bf{i}} + 6\mathop {\lim }\limits_{t \to \ln 2} \left( {{e^{ - t}}} \right){\bf{j}} - 4\mathop {\lim }\limits_{t \to \ln 2} \left( {{e^{ - 2t}}} \right){\bf{k}} \cr
& {\text{evalute each limit substitute }}\pi {\text{/2 for }}t \cr
& = 2\left( {{e^{\ln 2}}} \right){\bf{i}} + 6\left( {{e^{ - \ln 2}}} \right){\bf{j}} - 4\left( {{e^{ - 2\left( {\ln 2} \right)}}} \right){\bf{k}} \cr
& {\text{simplifying}} \cr
& = 2\left( 2 \right){\bf{i}} + 6\left( {\frac{1}{2}} \right){\bf{j}} - 4\left( {\frac{1}{4}} \right){\bf{k}} \cr
& = 4{\bf{i}} + 3{\bf{j}} - {\bf{k}} \cr} $$
