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

Options

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

Answer

To find the limit of the matrix \( A^n \) as \( n \) approaches infinity, where

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

we will follow these steps:

### Step 1: Calculate \( A^2 \)

To find \( A^2 \), we multiply \( A \) by itself:

\[ A^2 = A \cdot A = \begin{pmatrix} 1 & \frac{x}{n} \\ -\frac{x}{n} & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & \frac{x}{n} \\ -\frac{x}{n} & 1 \end{pmatrix}. \]

Calculating the elements:

- The (1,1) entry: \( 1 \cdot 1 + \frac{x}{n} \cdot -\frac{x}{n} = 1 - \frac{x^2}{n^2} \).
- The (1,2) entry: \( 1 \cdot \frac{x}{n} + \frac{x}{n} \cdot 1 = \frac{2x}{n} \).
- The (2,1) entry: \( -\frac{x}{n} \cdot 1 + 1 \cdot -\frac{x}{n} = -\frac{2x}{n} \).
- The (2,2) entry: \( -\frac{x}{n} \cdot \frac{x}{n} + 1 \cdot 1 = 1 - \frac{x^2}{n^2} \).

Thus, we have:

\[ A^2 = \begin{pmatrix} 1 - \frac{x^2}{n^2} & \frac{2x}{n} \\ -\frac{2x}{n} & 1 - \frac{x^2}{n^2} \end{pmatrix}. \]

### Step 2: Calculate \( A^3 \)

Now, we find \( A^3 = A^2 \cdot A \):

\[ A^3 = \begin{pmatrix} 1 - \frac{x^2}{n^2} & \frac{2x}{n} \\ -\frac{2x}{n} & 1 - \frac{x^2}{n^2} \end{pmatrix} \cdot \begin{pmatrix} 1 & \frac{x}{n} \\ -\frac{x}{n} & 1 \end{pmatrix}. \]

Calculating the elements:

- The (1,1) entry: \( (1 - \frac{x^2}{n^2}) \cdot 1 + \frac{2x}{n} \cdot -\frac{x}{n} = 1 - \frac{x^2}{n^2} - \frac{2x^2}{n^2} = 1 - \frac{3x^2}{n^2} \).
- The (1,2) entry: \( (1 - \frac{x^2}{n^2}) \cdot \frac{x}{n} + \frac{2x}{n} \cdot 1 = \frac{x}{n} - \frac{x^3}{n^3} + \frac{2x}{n} = \frac{3x}{n} - \frac{x^3}{n^3} \).
- The (2,1) entry: \( -\frac{2x}{n} \cdot 1 + (1 - \frac{x^2}{n^2}) \cdot -\frac{x}{n} = -\frac{2x}{n} - \frac{x}{n} + \frac{x^3}{n^3} = -\frac{3x}{n} + \frac{x^3}{n^3} \).
- The (2,2) entry: \( -\frac{2x}{n} \cdot \frac{x}{n} + (1 - \frac{x^2}{n^2}) \cdot 1 = -\frac{2x^2}{n^2} + 1 - \frac{x^2}{n^2} = 1 - \frac{3x^2}{n^2} \).

Thus, we have:

\[ A^3 = \begin{pmatrix} 1 - \frac{3x^2}{n^2} & \frac{3x}{n} - \frac{x^3}{n^3} \\ -\frac{3x}{n} + \frac{x^3}{n^3} & 1 - \frac{3x^2}{n^2} \end{pmatrix}. \]

### Step 3: Generalize \( A^n \)

From the patterns observed, we can generalize \( A^n \):

\[ A^n = \begin{pmatrix} 1 - \frac{nx^2}{n^2} & \frac{nx}{n} \\ -\frac{nx}{n} & 1 - \frac{nx^2}{n^2} \end{pmatrix} = \begin{pmatrix} 1 - \frac{3x^2}{n} & \frac{nx}{n} \\ -\frac{nx}{n} & 1 - \frac{3x^2}{n} \end{pmatrix}. \]

### Step 4: Take the Limit as \( n \to \infty \)

Now, we take the limit as \( n \) approaches infinity:

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

### Final Result

Thus, the limit is:

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

where \( I \) is the identity matrix.