Answer
$\langle-1,\ \displaystyle \frac{\pi}{2},0\rangle$
Work Step by Step
Component by component,
$\displaystyle \lim_{t\rightarrow\infty}\frac{1+t^{2}}{1-t^{2}}=\quad $divide numerator and denominator with $t^{2}$
$=\displaystyle \lim_{t\rightarrow\infty}\frac{(1/t^{2})+1}{(1/t^{2})-1}=\frac{0+1}{0-1}=\boxed{-1}$,
$\displaystyle \lim_{t\rightarrow\infty}\tan^{-1}t =\boxed{\displaystyle \frac{\pi}{2}} . $
$\displaystyle \lim_{t\rightarrow\infty}\frac{1-e^{-2t}}{t}=\lim_{t\rightarrow\infty}\frac{1}{t}-\frac{1}{te^{2t}}$
$=0-0=\boxed{0}$.
$\displaystyle \lim_{t\rightarrow\infty}\langle\frac{1+t^{2}}{1-t^{2}},\ \tan^{-1}t,\ \displaystyle \frac{1-e^{-2t}}{t}\rangle=\langle-1,\ \displaystyle \frac{\pi}{2},0\rangle$.
