# Chapter 11 - Infinite Series - 11.1 Sequences - Exercises - Page 537: 30

$0$

#### 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 $$\lim_{n\to \infty}\tan^{-1}{e^{-n}}= \tan^{-1}{ \lim_{n\to \infty}e^{-n}}=\tan^{-1}0=0.$$

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