Answer
$\lt -1,\ \dfrac{\pi}{2},0 \gt$ or, $ -i+\ \dfrac{\pi}{2}j$
Work Step by Step
$\lim\limits_{t \to \infty}\dfrac{1+t^{2}}{1-t^{2}}=\lim\limits_{t \to \infty}\dfrac{(1/t^{2})+1}{(1/t^{2})-1}=\lim\limits_{t \to \infty}\dfrac{(1/\infty)+1}{(1/\infty)-1}=\dfrac{0+1}{0-1}=-1$
and $\lim\limits_{t \to \infty} \tan^{-1}(t) =\tan^{-1}(\infty)=\dfrac{\pi}{2}$
Also, $\lim\limits_{t \to \infty} \dfrac{1-e^{-2t}}{t}=\lim\limits_{t \to \infty}\dfrac{1}{t}-\lim\limits_{t \to \infty} \dfrac{1}{te^{2t}}=0$
Hence, we have $\displaystyle \lim_{t\rightarrow\infty}\lt \dfrac{1+t^{2}}{1-t^{2}},\ \tan^{-1}t,\dfrac{1-e^{-2t}}{t} \gt=\lt -1,\ \dfrac{\pi}{2},0 \gt$ or, $ -i+\ \dfrac{\pi}{2}j$
