limit $\lim_{n\to ∞}\sin(\pi(2+\sqrt3)^n)$ $\lim_{n\to ∞}\sin(\pi(2+\sqrt3)^n)$
I tried to write it as $\sin (n\pi - \theta)$ to get the form $∞-∞$ form within $\sin$ function. But could not proceed after that. How should I do it?
Edit:I am sorry, I forgot to mention $n\in \mathbb{N}$
 A: Note $|2-\sqrt{3}|<1$ and hence $\lim_{n\to\infty}(2-\sqrt{3})^n=0$. 
Since $(2+\sqrt{3})^n+(2-\sqrt{3})^n$ is an integer, one has
\begin{eqnarray}
&&\lim_{n\to\infty}\sin[\pi(2+\sqrt{3})^n]\\
&=&\lim_{n\to\infty}\sin\bigg[\pi\big[(2+\sqrt{3})^n+(2-\sqrt{3})^n\big]-\pi(2-\sqrt{3})^n\bigg]\\
&=&\lim_{n\to\infty}\bigg\{\sin\bigg[\pi\big[(2+\sqrt{3})^n+(2-\sqrt{3})^n\big]\bigg]\cos\big[\pi(2-\sqrt{3})^n\big]\\
&&-\cos\bigg[\pi\big[(2+\sqrt{3})^n+(2-\sqrt{3})^n\big]\bigg]\sin\big[\pi(2-\sqrt{3})^n\big]\bigg\}\\
&=&0.
\end{eqnarray}
A: Hint(s): $(2+\sqrt{3})^n$ is closer and closer to an integer as $n$ increases, since
$$ (2+\sqrt{3})^n+(2-\sqrt{3})^n $$
is an integer and $|2-\sqrt{3}|<\frac{1}{3}$. The sine function is a Lipschitz-continuous function and $\sin(\pi m)=0$ for any integer $m$, hence your limit equals zero.
A: since 
$$
K(n)=(2+\sqrt3)^n+(2-\sqrt3)^n \in \mathbf{N}
$$
$$\lim_{n\to ∞}(2-\sqrt3)^n=0$$
You can do the following:
$$\lim_{n\to ∞}\sin(\pi(2+\sqrt3)^n)=\lim_{n\to ∞}\sin(\pi(K(n)-(2-\sqrt3)^n))=-\lim_{n\to ∞}\sin(\pi(2-\sqrt3)^n)=0
$$
