<s>[INST] Finding matrix $A^n,$ when $\lim_{n \to \infty}$ 
Finding $\lim_{n\rightarrow \infty}\begin{pmatrix}
1 & \frac{x}{n}\\ \\
-\frac{x}{n} & 1 
\end{pmatrix}^n$ for all $x\in \mathbb{R}$

Try:
 Let $$ A = \begin{pmatrix}1&\frac{x}{n}\\\\-\frac{x}{n}&1\end{pmatrix}.$$
Then $$ A^2 = \begin{pmatrix}1-\frac{x^2}{n^2}&\frac{2x}{n}\\\\-\frac{2x}{n}&1-\frac{x^2}{n^2}\end{pmatrix}$$
And then $$A^3 = \begin{pmatrix}1-3\frac{x^2}{n^2}&\frac{3x}{n}-\frac{x^3}{n^3}\\\\-3\frac{x}{n}+\frac{x^3}{n^3}&1-\frac{x^2}{n^2}\end{pmatrix}$$
So by using same way and taking $\lim_{n\rightarrow \infty}A^n = \begin{pmatrix}1&0\\\\0&1\end{pmatrix}$
But answer given as $$\begin{pmatrix}\cos x &\sin x\\\\-\sin x&\cos x\end{pmatrix}$$
Could some help me where I am missing and also explain how  to solve it?
 [/INST] We have
$$A = \sqrt{1 + \frac {x^2}{n^2}}\begin{pmatrix}\frac 1 {\sqrt{1 + \frac {x^2}{n^2}}}&\frac {\frac x n} {\sqrt{1 + \frac {x^2}{n^2}}}\\\\-\frac {\frac x n} {\sqrt{1 + \frac {x^2}{n^2}}}&\frac 1 {\sqrt{1 + \frac {x^2}{n^2}}}\end{pmatrix}$$
So if $\theta_n$ is such that $\cos \theta_n=\frac 1 {\sqrt{1 + \frac {x^2}{n^2}}}$ and $\sin \theta_n = -\frac {\frac x n} {\sqrt{1 + \frac {x^2}{n^2}}}$, and $\rho_n=\sqrt{1 + \frac {x^2}{n^2}}$, then
$$A=\rho_n\begin{pmatrix}\cos \theta_n&-\sin \theta_n\\\\\sin \theta_n&\cos \theta_n\end{pmatrix}$$
Therefore, $A$ is the product of the scaling factor $\rho_n$ with the rotation matrix of angle $\theta_n$. So powers of $A$ are 
$$A^n=\rho_n^n\begin{pmatrix}\cos n\theta_n&-\sin n\theta_n\\\\\sin n\theta_n&\cos n\theta_n\end{pmatrix}$$
Now, using Taylor expansions, $$\lim_{n\rightarrow +\infty}\rho_n^n=1$$ and $$\cos(n\theta_n)=\cos\left(n\arccos\bigg(\frac 1 {\sqrt{1 + \frac {x^2}{n^2}}}\bigg)\right)\rightarrow\cos(x)$$
$$\sin(n\theta_n)=\sin\left(n\arcsin\bigg(\frac {-\frac {x} n} {\sqrt{1 + \frac {x^2}{n^2}}}\bigg)\right)\rightarrow-\sin(x)$$
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