Answer
$0$
Work Step by Step
Using Theorem 1, we have
$$ \lim\limits_{n \to \infty}{e^{-n}}=0.$$
Now, since $e^{-x}$ is continuous, then by Theorem 4, we have
$$\lim_{n\to \infty}\tan^{-1}{e^{-n}}= \tan^{-1}{ \lim_{n\to \infty}e^{-n}}=\tan^{-1}0=0.$$
