Answer
$\langle 0,\ \displaystyle \frac{1}{2},1\rangle$
Work Step by Step
Component by component,
$\displaystyle \lim_{t\rightarrow\infty}te^{-t}=\lim_{t\rightarrow\infty}\frac{t}{e^{t}}\quad$[$\displaystyle \frac{0}{0}$ ... L'Hospital's Rule].
$=\displaystyle \lim_{t\rightarrow\infty}\frac{1}{e^{t}}=\boxed{0} $
$\displaystyle \lim_{t\rightarrow\infty}\frac{t^{3}+t}{2t^{3}-1}=\quad $... divide numerator and denominator with $t^{3}$
$\displaystyle \lim_{t\rightarrow\infty}\frac{1+(1/t^{2})}{2-(1/t^{3})}=\frac{1+0}{2-0}=\boxed{\frac{1}{2}}$
$\displaystyle \lim_{t\rightarrow\infty}t \displaystyle \sin\frac{1}{t}=\lim_{t\rightarrow\infty}\frac{\sin(1/t)}{1/t}=\quad\left[\begin{array}{lll}
u=1/t, & & \\
t\rightarrow\infty & \Rightarrow & u\rightarrow 0
\end{array}\right]$
$=\displaystyle \lim_{u\rightarrow 0}\frac{\sin u}{u}=\boxed{1}$
$\displaystyle \lim_{t\rightarrow\infty}\langle te^{-t},\ \displaystyle \frac{t^{3}+t}{2t^{3}-1},\ t\displaystyle \sin\frac{1}{t}\rangle=\langle 0,\ \displaystyle \frac{1}{2},1\rangle$
