Let `A={:[(1,x/n),(-x/n,1)]:},"then "lim_(ntooo) A^(n)` is:

Options

A. (a) `{:[(1,0),(0,1)]:}`
B. (b) `{:[(cosx,sinx),(-sinx,cosx)]:}`
C. (c) `{:[(cosx,-sinx),(sinx,cosx)]:}`
D. (d) `{:[(-1,0),(0,-1)]:}`

Answer

To solve the problem, we need to find the limit of the matrix \( A^n \) as \( n \) approaches infinity, where

\[ A = \begin{pmatrix} 1 & \frac{x}{n} \\ -\frac{x}{n} & 1 \end{pmatrix} \]

### Step 1: Evaluate the limit of matrix \( A \) as \( n \to \infty \)

We start by evaluating the limit of each element in the matrix \( A \):

\[ \lim_{n \to \infty} A = \begin{pmatrix} \lim_{n \to \infty} 1 & \lim_{n \to \infty} \frac{x}{n} \\ \lim_{n \to \infty} -\frac{x}{n} & \lim_{n \to \infty} 1 \end{pmatrix} \]

Calculating each limit:

- \( \lim_{n \to \infty} 1 = 1 \)
- \( \lim_{n \to \infty} \frac{x}{n} = 0 \)
- \( \lim_{n \to \infty} -\frac{x}{n} = 0 \)

Thus, we have:

\[ \lim_{n \to \infty} A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \]

where \( I \) is the identity matrix.

### Step 2: Determine \( A^n \) as \( n \to \infty \)

Since we found that \( \lim_{n \to \infty} A = I \), we can analyze \( A^n \):

\[ \lim_{n \to \infty} A^n = \lim_{n \to \infty} I^n \]

The identity matrix raised to any power is still the identity matrix:

\[ I^n = I \]

### Step 3: Conclude the limit

Therefore, we conclude:

\[ \lim_{n \to \infty} A^n = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Final Answer

The limit as \( n \to \infty \) of \( A^n \) is the identity matrix:

\[ \boxed{\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}} \]