How find this limit $I=\lim_{n\to\infty}(1+\sin{(\sqrt{4n^2+1}\cdot\pi)})^n$ find this limit
$$I=\lim_{n\to\infty}(1+\sin{(\sqrt{4n^2+1}\cdot\pi)})^n$$
My try: let 
$$I=e^{\lim_{n\to\infty}n\sin{(\sqrt{4n^2+1}\pi})}$$
 A: $$\sin((\sqrt{4n^2+1})\pi)=\sin(2\pi n(\sqrt{1+\frac{1}{4n^2}}))=\sin(2\pi n+2 \pi n\frac{1}{8n^2}+O(1/n^3))$$
$$=\sin(\frac{\pi}{4n}+O(1/n^3))=\frac{\pi}{4n}+O(\frac{1}{n^3})$$
A: My solution:
$$I=e^{\displaystyle\lim_{n\to\infty}n\sin{(\sqrt{4n^2+1}\pi)}}=e^{\displaystyle\lim_{n\to\infty}n\sin{(\sqrt{4n^2+1}-2n)\pi}}$$
then
$$\lim_{n\to\infty}n\sin{\dfrac{1}{\sqrt{4n^2+1}+2n}\pi}=\lim_{n\to\infty}\dfrac{n}{\sqrt{4n^2+1}+2n}=\dfrac{\pi}{4}$$
so
$$I=\lim_{n\to\infty}(1+\sin{(\sqrt{4n^2+1}\pi)})^n=e^{\frac{\pi}{4}}$$
A: So, you would like to know how $x_n=n\sin(z_n\pi)$ with $z_n=\sqrt{4n^2+1}$ behaves when $n\to\infty$. The thing to realize is that $z_n$ is almost $2n$ hence $\sin{(\sqrt{4n^2+1}\pi})$ is close to $\sin(2n\pi)=0$. How close?
A first method is to note that $z_n=2n\sqrt{1+1/(4n^2)}$ and that, when $u\to0$, $\sqrt{1+u}=1+u/2+o(u)$. 
Hence $z_n=2n(1+1/(8n^2)+o(1/n^2))=2n+1/(4n)+o(1/n)$ and $\sin(z_n\pi)=\sin(\pi/(4n)+o(1/n))=\pi/(4n)+o(1/n)$, which yields $x_n=\pi/4+o(1)$, that is,
$$\lim_{n\to\infty}x_n=\pi/4.
$$
A second method is to compute directly $z_n-2n$, noting that 
$$
z_n-2n=\frac{z_n^2-4n^2}{z_n+2n}=\frac1{z_n+2n}.
$$
Since $z_n\sim2n$, this yields directly $z_n-2n\sim1/4n$ and the same conclusion follows.
