Answer
$$\lim _{t \rightarrow 0}\left[e^{t} \mathbf{i}+\frac{\sin t}{t} \mathbf{j}+e^{-t} \mathbf{k}\right]=\mathbf{i}+\mathbf{j}+\mathbf{k}$$
Work Step by Step
Given $$\lim _{t \rightarrow 0}\left[e^{t} \mathbf{i}+\frac{\sin t}{t} \mathbf{j}+e^{-t} \mathbf{k}\right]$$
Since
\begin{align}
&\lim _{t \rightarrow 0} e^t =1\\
&\lim _{t \rightarrow 0} e^{-t} =1\\
&\lim _{t \rightarrow 0} \frac{\sin t}{t} =\lim _{t \rightarrow 1} \frac{ \cos t}{1} =\cos 0=1\ \quad (\text{L'Hôpital Rule})\\
\end{align}
So, we get
\begin{align}
L&=\lim _{t \rightarrow 0}\left[e^{t} \mathbf{i}+\frac{\sin t}{t} \mathbf{j}+e^{-t} \mathbf{k}\right]\\
&=\lim _{t \rightarrow 0} e^{t} \mathbf{i}+\lim _{t \rightarrow 0}\frac{\sin t}{t} \mathbf{j}+\lim _{t \rightarrow 0}e^{-t} \mathbf{k} \\
&=\mathbf{i}+\mathbf{j}+\mathbf{k}
\end{align}
