Given $A=\left[\begin{smallmatrix}\cos\theta&\sin \theta\\-\sin\theta&\cos\theta\end{smallmatrix}\right]$, find $\lim_{n \to \infty}\frac {A^n}{n} $ $A^n =\left[\begin{matrix} \cos n\theta & \sin n\theta\\-\sin n\theta & \cos n\theta  \end{matrix}\right] $
Now $\theta \in \mathbb{R}$
This means 
$$\lim_{n \to \infty}\dfrac {A^n}{n} = \lim_{n \to \infty}\left[\begin{matrix} \dfrac{\cos n\theta}{n} & \dfrac{\sin n\theta}{n}\\\dfrac{-\sin n\theta}{n} & \dfrac{\cos n\theta}{n}  \end{matrix}\right]$$ 
Now I think this should be $= \left[\begin{matrix} 0 & 0\\0 & 0  \end{matrix}\right]$
But I do not understand whether this is a correct conclusion or not.
Sorry its easy so very very easy . Expand $\cos n\theta$ using Taylor Maclaurin series and divide by n
$\implies$ 0 
So is true for $ \sin n\theta$
Thus the answer indeed is $= \left[\begin{matrix} 0 & 0\\0 & 0  \end{matrix}\right]$
 A: The matrix $A=\begin{bmatrix}\cos(\theta)&\sin(\theta)\\-\sin(\theta)&\cos(\theta)\end{bmatrix}$ is a particularly important example and represents a clockwise rotation transformation by an angle of $\theta$.
You have correctly calculated the result, identified, or remembered the property that having repeatedly applied a rotation of $\theta$ some $n$ times successively that this is the same result as having gone through a single rotation of an angle of $n\theta$.  That is to say $A^n = \begin{bmatrix}\cos(n\theta)&\sin(n\theta)\\-\sin(n\theta)&\cos(n\theta)\end{bmatrix}$.
Next, you correctly noted that $\dfrac{A^n}{n}=\begin{bmatrix}\frac{\cos(n\theta)}{n}&\frac{\sin(n\theta)}{n}\\ -\frac{\sin(n\theta)}{n}&\frac{\cos(n\theta)}{n}\end{bmatrix}$
From here, to compute $\lim\limits_{n\to\infty}\dfrac{A^n}{n}$, it suffices to compute the limits of each individual entry.
Note that regardless the value of $n$ and regardless the value of $\theta$ we have $-1\leq \cos(n\theta)\leq 1$.  Similarly for $\sin(n\theta)$ and $-\sin(n\theta)$.
As a result since for all values of $n$ and all values of $\theta$ we have $\dfrac{-1}{n}\leq \dfrac{\cos(n\theta)}{n}\leq \dfrac{1}{n}$ and both of the outer limits both approach zero as $n$ grows large, so too must the limit of the center expresion approach zero as $n$ grows large as per the squeeze theorem.  The proofs for the other entries are identical.
It follows then that $\lim\limits_{n\to\infty}\dfrac{A^n}{n}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$, as you expected.
