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 
$$
\lim_{n\to\infty} \sin\bigl(\pi\sqrt{n^2+1}\bigr)=0 
$$

$$\lim_{n\to\infty}\sin\pi\sqrt{n^2+1}=\lim_{n\to\infty}(-1)^n\sin(\sqrt{n^2+1}-n)\pi$$ $$=\lim_{n\to\infty}(-1)^n\sin(\frac{1}{\sqrt{n^2+1}+n})\pi=\lim_{n\to\infty}(-1)^n\sin0\cdot\pi$$ $$=0$$
