Answer
\[\mathbf{i}\]
Work Step by Step
\[\begin{align}
& \underset{t\to 0}{\mathop{\lim }}\,\left( \frac{\sin t}{t}\mathbf{i}-\frac{{{e}^{t}}-t-1}{t}\mathbf{j}+\frac{\cos t+{{t}^{2}}/2-1}{{{t}^{2}}}\mathbf{k} \right) \\
& \text{Evaluate the limit} \\
& =\underset{t\to 0}{\mathop{\lim }}\,\frac{\sin t}{t}\mathbf{i}-\underset{t\to 0}{\mathop{\lim }}\,\frac{{{e}^{t}}-t-1}{t}\mathbf{j}+\underset{t\to 0}{\mathop{\lim }}\,\frac{\cos t+{{t}^{2}}/2-1}{{{t}^{2}}}\mathbf{k} \\
& =\left( 1 \right)\mathbf{i}-\frac{{{e}^{0}}-0-1}{0}\mathbf{j}+\frac{\cos 0+{{0}^{2}}/2-1}{{{0}^{2}}}\mathbf{k} \\
& =\mathbf{i}-\frac{0}{0}\mathbf{j}+\frac{0}{0}\mathbf{k} \\
& \text{Apply the L }\!\!'\!\!\text{ Hopitals Rule} \\
& \underset{t\to 0}{\mathop{\lim }}\,\frac{{{e}^{t}}-t-1}{t}=\underset{t\to 0}{\mathop{\lim }}\,\left( {{e}^{t}}-1 \right)=0 \\
& \underset{t\to 0}{\mathop{\lim }}\,\frac{\cos t+{{t}^{2}}/2-1}{{{t}^{2}}}=\underset{t\to 0}{\mathop{\lim }}\,\frac{-\sin t+t}{2t}=\underset{t\to 0}{\mathop{\lim }}\,\frac{-\cos t+1}{2}=\frac{0}{2}=0 \\
& \text{Therefore,} \\
& =\underset{t\to 0}{\mathop{\lim }}\,\frac{\sin t}{t}\mathbf{i}-\underset{t\to 0}{\mathop{\lim }}\,\frac{{{e}^{t}}-t-1}{t}\mathbf{j}+\underset{t\to 0}{\mathop{\lim }}\,\frac{\cos t+{{t}^{2}}/2-1}{{{t}^{2}}}\mathbf{k} \\
& =\left( 1 \right)\mathbf{i}-\left( 0 \right)\mathbf{j}+\left( 0 \right)\mathbf{k} \\
& =\mathbf{i} \\
\end{align}\]
