Show that $\,\,\, \lim_{n\to\infty} \sin\bigl(\pi\sqrt{n^2+1}\bigr)=0 $ Can anyone help me to solve this problem?
Show that 
$$
\lim_{n\to\infty} \sin\bigl(\pi\sqrt{n^2+1}\bigr)=0 
$$
 A: We have that
$$
0<\delta_n=\sqrt{n^2+1}-n=\frac{\big(\sqrt{n^2+1}+n\big)\big(\sqrt{n^2+1}-n\big)}{\sqrt{n^2+1}+n}=
\frac{1}{\sqrt{n^2+1}+n}<\frac{1}{2n}
$$
So 
$$
\sin\left(\pi\sqrt{n^2+1}\right)=\sin\left(\pi\sqrt{n^2+1}-\pi n+\pi n\right)=\sin(\pi\delta_n+\pi n)=(-1)^n\sin(\pi\delta_n),
$$
and as $\lvert \sin x\rvert\le \lvert x\rvert$, then
$$
\big|\sin\left(\pi\sqrt{n^2+1}\right)\big|=\lvert\sin(\pi\delta_n)\rvert\le \lvert\pi \delta_n\rvert<\frac{\pi}{2n}\to 0.
$$
Note. In fact, one can use the above to show that
$$
\lim_{n\to\infty} (-1)^n n\sin\big(\pi\sqrt{n^2+1}\big)=\frac{\pi}{2}.
$$
A: We have by the Taylor series:
$$\sqrt{n^2+1}=n\sqrt{1+\frac1{n^2}}=n\left(1+O\left(\frac1{n^2}\right)\right)=n+O\left(\frac1{n}\right)$$
hence
$$\sin\left(\pi\sqrt{n^2+1}\right)=\sin\left(n\pi+O\left(\frac1{n}\right)\right)=(-1)^n\sin\left(O\left(\frac1{n}\right)\right)\xrightarrow{n\to\infty}\ 0$$
A: Assuming $n$ to be integer,
$$\sin\left(\pi\sqrt{n^2+1}\right)=(-1)^{n-1}\sin\left(n\pi-\pi\sqrt{n^2+1}\right)$$
$$=(-1)^{n-1}\sin\left[\pi(n-\sqrt{n^2+1})\right]$$
$$=(-1)^{n-1}\sin\left[\pi\left(\frac{n^2-(n^2+1)}{n+\sqrt{n^2+1}}\right)\right]$$
$$=(-1)^{n-1}\sin\left[-\pi\left(\frac1{n+\sqrt{n^2+1}}\right)\right]$$
Now $\displaystyle\lim_{n\to\infty}\left(\frac1{n+\sqrt{n^2+1}}\right)=0$
A: $$\lim_{n\to\infty}\sin\pi\sqrt{n^2+1}=\lim_{n\to\infty}(-1)^n\sin(\sqrt{n^2+1}-n)\pi$$ $$=\lim_{n\to\infty}(-1)^n\sin(\frac{1}{\sqrt{n^2+1}+n})\pi=\lim_{n\to\infty}(-1)^n\sin0\cdot\pi$$ $$=0$$
