Answer
\[\mathbf{i}-3\mathbf{j}+\mathbf{k}\]
Work Step by Step
\[\begin{align}
& \underset{t\to 0}{\mathop{\lim }}\,\left( \frac{\tan t}{t}\mathbf{i}-\frac{3t}{\sin t}\mathbf{j}+\sqrt{t+1}\mathbf{k} \right) \\
& \text{Evaluate the limit} \\
& =\underset{t\to 0}{\mathop{\lim }}\,\frac{\tan t}{t}\mathbf{i}-\underset{t\to 0}{\mathop{\lim }}\,\frac{3t}{\sin t}\mathbf{j}+\underset{t\to 0}{\mathop{\lim }}\,\sqrt{t+1}\mathbf{k} \\
& \text{Apply the L }\!\!'\!\!\text{ Hopitals Rule} \\
& \underset{t\to 0}{\mathop{\lim }}\,\frac{\tan t}{t}=\underset{t\to 0}{\mathop{\lim }}\,\frac{{{\sec }^{2}}t}{1}={{\sec }^{2}}0=1 \\
& \underset{t\to 0}{\mathop{\lim }}\,\frac{3t}{\sin t}=\underset{t\to 0}{\mathop{\lim }}\,\frac{3}{\cos t}=\frac{3}{\cos 0}=3 \\
& \text{Therefore,} \\
& =\underset{t\to 0}{\mathop{\lim }}\,\frac{\tan t}{t}\mathbf{i}-\underset{t\to 0}{\mathop{\lim }}\,\frac{3t}{\sin t}\mathbf{j}+\underset{t\to 0}{\mathop{\lim }}\,\sqrt{t+1}\mathbf{k} \\
& =\left( 1 \right)\mathbf{i}-\left( 3 \right)\mathbf{j}+\sqrt{0+1}\mathbf{k} \\
& =\mathbf{i}-3\mathbf{j}+\mathbf{k} \\
\end{align}\]
