Answer
\begin{align}
\lim _{t \rightarrow 0}\left[t^{2} \mathbf{i}+3 t \mathbf{j}+\frac{1-\cos t}{t} \mathbf{k}\right]=0
\end{align}
Work Step by Step
Given $$\lim _{t \rightarrow 0}\left[t^{2} \mathbf{i}+3 t \mathbf{j}+\frac{1-\cos t}{t} \mathbf{k}\right]=0$$
Since
\begin{align}
\lim _{t \rightarrow 0} \frac{1-\cos t}{t} = \lim _{t \rightarrow 0} \frac{\sin t}{1}=\sin 0=0 \quad \text{ (L'Hôpital's Rule)}
\end{align}
So, we get
\begin{align}
L&=\lim _{t \rightarrow 0}\left[t^{2} \mathbf{i}+3 t \mathbf{j}+\frac{1-\cos t}{t} \mathbf{k}\right]\\
&= \lim _{t \rightarrow 0} t^{2} \mathbf{i}+\lim _{t \rightarrow 0}3 t \mathbf{j}+\lim _{t \rightarrow 0}\frac{1-\cos t}{t} \mathbf{k} \\
&= 0 \mathbf{i}+0 \mathbf{j}+0\mathbf{k}=0
\end{align}
