How to calculate the limit $\lim_{n \to \infty}n\sin(2\pi\sqrt{1+n^2})$? The limit is: $$\lim_{n \to \infty}n\sin(2\pi\sqrt{1+n^2}),n\in \mathbb N$$
I wasn't able to convert it to any of the standard indeterminate forms so I thought of doing it analytically.
I realized that $\sin$ would only yield bounded real values which are periodic. This suggests some kind of oscillatory behaviour but I'm not sure how this could help me evaluate this limit. 
Any hints or alternate approach?  

Firt note that


*

*$\sin 2\pi \sqrt{1+n^2} = \sin \left(2\pi \sqrt{1+n^2}- 2\pi n + 2\pi n\right) = \sin \frac{2\pi}{\sqrt{1+n^2} + n}$
Now, use $\lim_{x\to 0}\frac{\sin x}{x} = 1$ as follows
\begin{eqnarray*} n \sin \frac{2\pi}{\sqrt{1+n^2} + n}
& = & \frac{\sin \frac{2\pi}{\sqrt{1+n^2} + n}
}{\frac{2\pi}{\sqrt{1+n^2} + n}}\cdot \underbrace{\frac{2\pi n }{\sqrt{1+n^2} + n}
}_{\stackrel{n\to\infty}{\longrightarrow}\pi}\\
& \stackrel{n\to\infty}{\longrightarrow} & \pi
\end{eqnarray*}
