limit $\lim_{n\to ∞}\sin(\pi(2+\sqrt3)^n)$

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}

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.

since $$ K(n)=(2+\sqrt3)^n+(2-\sqrt3)^n \in \mathbf{N} $$ $$\lim_{n\to ∞}(2-\sqrt3)^n=0$$ You can do the following: $$\lim_{n\to ∞}\sin(\pi(2+\sqrt3)^n)=\lim_{n\to ∞}\sin(\pi(K(n)-(2-\sqrt3)^n))=-\lim_{n\to ∞}\sin(\pi(2-\sqrt3)^n)=0 $$