Let `y_(n)(x) = x^(2) + (x^(2))/(1+x^(2))+(x^(2))/((1+x^(2))^(2))+......(x^(2))/((1+x^(2))^(n-1))and y(x) = lim_(n rarr oo) y_(n) (x)`. Discuss the continuity of `y_(n)(x)(n = 1, 2, 3....n) and y(x) "at x" = 0`

Answer

To solve the problem, we will first analyze the function \( y_n(x) \) and then find the limit as \( n \) approaches infinity to determine \( y(x) \). Finally, we will discuss the continuity of both \( y_n(x) \) and \( y(x) \) at \( x = 0 \).

### Step 1: Write down the expression for \( y_n(x) \)

The function is given as:

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

### Step 2: Identify the series as a geometric progression (GP)

Notice that the terms after the first can be expressed as a geometric series. The first term \( a = x^2 \) and the common ratio \( r = \frac{1}{1+x^2} \).

### Step 3: Use the formula for the sum of a geometric series

The sum of the first \( n \) terms of a geometric series can be calculated using the formula:

\[ S_n = \frac{a(1 - r^n)}{1 - r} \]

Substituting \( a = x^2 \) and \( r = \frac{1}{1+x^2} \):

\[ y_n(x) = \frac{x^2 \left(1 - \left(\frac{1}{1+x^2}\right)^{n}\right)}{1 - \frac{1}{1+x^2}} \]

### Step 4: Simplify the expression

The denominator simplifies to:

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

Thus, we have:

\[ y_n(x) = \frac{x^2 \left(1 - \left(\frac{1}{1+x^2}\right)^{n}\right)}{\frac{x^2}{1+x^2}} = (1+x^2) \left(1 - \left(\frac{1}{1+x^2}\right)^{n}\right) \]

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

Now, we need to find \( y(x) = \lim_{n \to \infty} y_n(x) \):

\[ y(x) = (1+x^2) \left(1 - \lim_{n \to \infty} \left(\frac{1}{1+x^2}\right)^{n}\right) \]

Since \( \left(\frac{1}{1+x^2}\right)^{n} \) approaches 0 as \( n \to \infty \) for \( x \neq 0 \):

\[ y(x) = (1+x^2)(1 - 0) = 1 + x^2 \]

### Step 6: Analyze continuity at \( x = 0 \)

1. **For \( y_n(x) \)**:
- \( y_n(0) = 0^2 + \frac{0^2}{1+0^2} + \ldots + \frac{0^2}{(1+0^2)^{n-1}} = 0 \)
- Since \( y_n(x) \) is a polynomial function, it is continuous for all \( x \), including \( x = 0 \).

2. **For \( y(x) \)**:
- \( y(0) = 1 + 0^2 = 1 \)
- \( y(x) \) is also a polynomial function, hence continuous for all \( x \).

### Conclusion

- \( y_n(x) \) is continuous at \( x = 0 \) for all \( n \).
- \( y(x) \) is continuous at \( x = 0 \).