For each positive integer `n gt a,A_(n)` represents the area of the region restricted to the following two inequalities : `(x ^(2))/(n ^(2)) + y ^(2) and x ^(2)+ (y ^(2))/(n ^(2)) lt 1.` Find `lim _(n to oo) A_(n).`

Options

A. 4
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to find the limit of the area \( A_n \) defined by the inequalities:

1. \(\frac{x^2}{n^2} + y^2 < 1\)
2. \(x^2 + \frac{y^2}{n^2} < 1\)

as \( n \) approaches infinity.

### Step 1: Understand the inequalities

The inequalities describe two ellipses:

- The first inequality \(\frac{x^2}{n^2} + y^2 < 1\) describes an ellipse centered at the origin with semi-major axis \( n \) along the x-axis and semi-minor axis \( 1 \) along the y-axis.
- The second inequality \(x^2 + \frac{y^2}{n^2} < 1\) describes an ellipse centered at the origin with semi-major axis \( 1 \) along the x-axis and semi-minor axis \( n \) along the y-axis.

### Step 2: Find the intersection points

To find the area \( A_n \), we need to determine the intersection points of these two ellipses. Setting the equations equal to each other:

1. From the first equation, we have \( y^2 = 1 - \frac{x^2}{n^2} \).
2. From the second equation, we have \( y^2 = n^2(1 - x^2) \).

Setting these equal gives:

\[ 1 - \frac{x^2}{n^2} = n^2(1 - x^2) \]

### Step 3: Solve for \( x \)

Rearranging the equation, we find:

\[ 1 - n^2 + n^2 x^2 = \frac{x^2}{n^2} \]

Multiplying through by \( n^2 \) to eliminate the fraction:

\[ n^2 - n^4 + n^4 x^2 = x^2 \]

This simplifies to:

\[ (n^2 - n^4) = (1 - n^4)x^2 \]

From this, we can find \( x \):

\[ x^2 = \frac{n^2 - n^4}{1 - n^4} \]

### Step 4: Calculate the area \( A_n \)

The area \( A_n \) can be computed using integration. The area can be found by integrating the upper half of the first ellipse and subtracting the area under the second ellipse.

The area of the first ellipse can be computed as:

\[ A_1 = \int_{0}^{n} \sqrt{1 - \frac{x^2}{n^2}} \, dx \]

The area of the second ellipse can be computed as:

\[ A_2 = \int_{0}^{1} \sqrt{n^2(1 - x^2)} \, dx \]

### Step 5: Find the limit as \( n \to \infty \)

As \( n \) approaches infinity, the area \( A_n \) approaches the area of the unit circle, which is \( \pi \). Thus:

\[ \lim_{n \to \infty} A_n = \pi \]

### Final Answer

\[ \lim_{n \to \infty} A_n = \pi \]