Evaluating $\lim\limits_{n\to \infty} \sin \left( (2 + \sqrt 3 )^n\pi\right)$ for $n \in \mathbb N$ 
Evalulate $$\lim\limits_{n\to \infty} \sin \bigl( (2 + \sqrt 3 )^n\pi \bigr) \quad \text{ for } n \in \mathbb N$$

This question appeared in my high school exam.
My first idea was as $n$ is an integer then the value must change abruptly for every increase in $n$ which makes the function discontinuous and so the limit does not exist.
But there is another method that proves the limit to be $0$.
Solution :
$$
(2+\sqrt{3})^n+(2-\sqrt{3})^n=2m,m\in I^*
$$
\begin{align}
\therefore \lim_{n\to\infty} \sin\Big((2+\sqrt{3})^n\pi\Big)
&=\lim_{n\to \infty} \sin\Big((2m-(2-\sqrt{3})^n)\pi\Big)\\
&=\lim_{n\to \infty} \sin\Big(2m\pi-(2-\sqrt{3})^n\pi\Big)\\
&=-\lim_{n\to \infty} \sin\Big((2-\sqrt{3})^n\pi\Big)\\
&=-\lim_{n\to \infty} \sin\Bigg(\frac{\pi}{(2+\sqrt{3})^n}\Bigg)\\
&=0
\end{align}
(original images 1 and 2)
So, where did I go wrong.

Note that 
$$\lim_{n\to\infty}\sin\left((2+\sqrt 3)^n \right) $$
is not the same as
$$\lim_{n\to\infty}\sin\left((2+\sqrt 3)^n \pi\right) $$
