Answer
converges to $1$
Work Step by Step
As we know that when $x \gt 0$ so, $\lim\limits_{n \to \infty}e^n=\infty$
Let $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \tan hn $
This implies that $ \lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \tan hn =\lim\limits_{n \to \infty} \dfrac{e^n-e^{-n}}{e^n+e^{-n}}$
$\lim\limits_{n \to \infty} \dfrac{1-(1/e^{2n})}{1 +(1/e^{2n})} =1$
Thus, $\lim\limits_{n \to \infty} a_n=1$ and {$a_n$} converges to $1$.
