# recursive matrix with infinite limit

If $$A=\begin{bmatrix} 1 & -\frac{x}{n}\\ \\ \frac{x}{n} & 1 \end{bmatrix}$$, then $$\displaystyle \lim_{x\rightarrow 0}\left(\lim_{n\rightarrow \infty}\frac{1}{x}(I-A^n)\right)=\,?$$

what I tried:

$$A^2=\begin{bmatrix} 1 & -\frac{x}{n}\\ \\ \frac{x}{n} & 1 \end{bmatrix}\begin{bmatrix} 1 & -\frac{x}{n}\\ \\ \frac{x}{n} & 1 \end{bmatrix}=\begin{bmatrix} 1-\frac{x}{n^2} & -\frac{2x}{n}\\ \\ -\frac{2x}{n} & 1-\frac{x^2}{n^2} \end{bmatrix}$$

$$A^3=\begin{bmatrix} 1-\frac{x}{n^2} & -\frac{2x}{n}\\ \\ -\frac{2x}{n} & 1-\frac{x^2}{n^2} \end{bmatrix}\begin{bmatrix} 1 & -\frac{x}{n}\\ \\ \frac{x}{n} & 1 \end{bmatrix}=\begin{bmatrix} 1-\frac{3x}{n^2} & \frac{2x}{n}-\frac{x^3}{n^3}\\ \\ \frac{3x}{n}-\frac{x^3}{n^3} & 1-\frac{3x^2}{n^2} \end{bmatrix}$$

but for finding $$A^4$$ is very hard

How do I solve it? Help me please.

• Use similarity transform to diagonalize matrix $A$. – 87091403130 Feb 17 '20 at 15:07
• You have a typo in $A^2$ in the upper left element: It's $1-x^2/n^2$, not $1-x/n^2$. Maybe that will simplify matters? Moreover abbreviating $a=x/n$ or so might save you some writing. – emacs drives me nuts Feb 17 '20 at 15:48

$$A=I+(x/n) B$$ where, $$B=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$ and $$B^2=-I, B^3=-B, B^4=I, B^5=B,....$$ Next, by binomial expansion $$A^n=[I+(x/n)B)]^n =I+{n \choose 1} (x/n) B+ {n \choose 2}(x/n)^2 (-I)+ {n \choose 3}(x/n)^3 (-B)+{n\choose 4} (x/n)^4 (I)+....+(x/n)^n B^n.$$ $$\implies A^n=\left[1-{n \choose 2}(x/n)^2+{n \choose 4}(x/n)^4 -...\right]I+\left[(x/n)-{n \choose 3}(x/n)^3+{n \choose 5}(x/n)^5-...\right] B.$$ $$\implies I- A^n = I-\frac{1}{2} ( [(1+(x/n))^n + (1-(x/n))^n] I + [(1+(x/n))^n-(1-(x/n))^n] B ).$$ $$\implies \lim_{x \rightarrow 0} \lim_{n \rightarrow \infty}\frac{I-A^n}{x} = \lim_{x\rightarrow 0} \left(I-\frac{1}{2x}([e^x+e^{-x}] I+ [e^x-e^{-x}] B \right)$$ $$=\lim_{x \rightarrow 0}\begin{pmatrix} \frac{1-\cosh x}{x}& -\frac{\sinh x}{x}) \\ \frac{\sinh x}{x} & \frac{1-\cosh x}{x} \end {pmatrix}=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$