If `f(x)=x^(2)+(x^(2))/(1+x^(2))+(x^(2))/((1+x^(2))^(2))+……` upto, `oo`, then

Options

A. `lim_(xrarr0)f(x)` does not exist
B. `f(x)` is continuous but not differentiable at x = 0
C. f(x) is discontinuous at x = 0
D. f(x) is differentiable at x = 0

Answer

To solve the problem, we need to analyze the function given by the series:

\[ f(x) = x^2 + \frac{x^2}{1+x^2} + \frac{x^2}{(1+x^2)^2} + \ldots \]

This series can be recognized as a geometric series. Let's break it down step by step.

### Step 1: Identify the first term and common ratio
The first term \( a \) of the series is \( x^2 \). The subsequent terms can be expressed in terms of the first term and the common ratio \( r \).

The common ratio \( r \) can be identified as:
\[ r = \frac{1}{1+x^2} \]

### Step 2: Write the sum of the infinite geometric series
The sum \( S \) of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
where \( |r| < 1 \).

In our case:
- First term \( a = x^2 \)
- Common ratio \( r = \frac{1}{1+x^2} \)

### Step 3: Apply the formula
Now we can substitute \( a \) and \( r \) into the formula:
\[ f(x) = \frac{x^2}{1 - \frac{1}{1+x^2}} \]

### Step 4: Simplify the expression
Now we simplify the denominator:
\[ 1 - \frac{1}{1+x^2} = \frac{(1+x^2) - 1}{1+x^2} = \frac{x^2}{1+x^2} \]

So we have:
\[ f(x) = \frac{x^2}{\frac{x^2}{1+x^2}} = x^2 \cdot \frac{1+x^2}{x^2} = 1 + x^2 \]

### Step 5: Conclusion
Thus, the function simplifies to:
\[ f(x) = 1 + x^2 \]