## Thomas' Calculus 13th Edition

converges to $1$
Let $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \sqrt n \sin (\dfrac{1}{\sqrt n})$ But $\lim\limits_{n \to \infty} \sqrt n \sin (\dfrac{1}{\sqrt n})=\dfrac{0}{0}$ This shows that the limit has an Indeterminate form, so we will use L-Hospital's rule. This implies that $\lim\limits_{n \to \infty} \dfrac{\cos(\dfrac{1}{\sqrt n}) (\dfrac{-1}{2n^{3/2}})}{(1/2n^{3/2})}=\lim\limits_{n \to \infty} \cos (1/\sqrt n)$ and $\lim\limits_{n \to \infty} \cos (\dfrac{1}{\sqrt n})=1$ Hence, $\lim\limits_{n \to \infty} a_n=1$ and {$a_n$} converges to $1$.