Answer
$\dfrac{\pi}{2}$
Work Step by Step
Let us consider that $\lim\limits_{x \to \infty}a_n=\lim\limits_{x \to \infty} \tan^{-1} x\\=\tan^{-1} (\infty) \\=\dfrac{\pi}{2}$
Since, $\tan (\dfrac{\pi}{2})=\infty$
Thus, $\lim\limits_{x \to \infty}a_n=\dfrac{\pi}{2}$
