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?  
 A: Use the binomial approximation  for large $n$ as $$\sqrt{1+n^2}=n \sqrt{1+1/n^2} \sim n (1+1/(2n^2)) \sim n+1/(2n).$$ Then
$$L=\lim_{n\rightarrow \infty} n\sin(2\pi\sqrt{1+n^2})= \lim_{n \rightarrow \infty} n \sin(2\pi n+ \pi/n)= \lim_{n\rightarrow \infty} n \sin (\pi/n) =\lim_{n \rightarrow \infty} n (\pi/n)= \pi. $$
A: 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*}
A: Hint:
$$\sin\left(2\pi\sqrt{1+n^2}\right)=\sin\left(2\pi\sqrt{1+n^2}-2\pi n\right)=\sin\left(\frac{2\pi}{\sqrt{1+n^2}+n}\right)\sim \frac{2\pi}{2n}.$$
Though the continuous function oscillates with an increasing amplitude, the sampled values do converge:

A: $\lim_{n\to\infty}(-nSin(2nπ-2π\sqrt{n^2+1})$ $\\$
=$\lim_{n\to\infty}(nSin(\frac{(2π)}{\sqrt{(n^2+1)}+n} )$. $\\$
=$\lim_{n\to\infty}(2π)(\frac{n}{\sqrt{(n^2+1)}+n})$. $\\$
=$\frac{2π}{2}$=π
A: As in other answers and comments
$$y=\sin(2\pi\sqrt{1+n^2})=\sin(2\pi\sqrt{1+n^2}-2n \pi)$$
Now, by composition of Taylor expansions
$$\sqrt{1+n^2}-n=\frac{1}{2 n}-\frac{1}{8 n^3}+O\left(\frac{1}{n^5}\right)$$
$$y=\sin\left(\frac{\pi }{n}-\frac{\pi }{4 n^3} +O\left(\frac{1}{n^5}\right)\right)=\frac{\pi }{n}-\frac{\pi  \left(3+2 \pi ^2\right)}{12
   n^3}+O\left(\frac{1}{n^5}\right)$$ making
$$n y=\pi -\frac{\pi  \left(3+2 \pi ^2\right)}{12 n^2}+O\left(\frac{1}{n^4}\right)$$ which shows the limit and also how it is approached.
Using $n=10$, the "exact" value would be $3.08274$ while the above asymptotics gives $\frac{1197 \pi -2 \pi ^3}{1200} \approx  3.08206$
