If `f(x)=x^2+(x^2)/(1+x^2)+(x^2)/((1+x^2)^2)+ ....oo` term then at `x=0,f(x)`

Options

A. `lim_(x to 0) 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 find the value of the function \( f(x) \) defined as:

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

This series can be recognized as an infinite geometric series.

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

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

### Step 2: 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 \( |r| < 1 \).

### Step 3: Substitute the values of \( a \) and \( r \)
Substituting the values of \( a \) and \( r \):

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

### Step 4: Simplify the expression
Now, simplify the denominator:

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

Thus, we can rewrite \( f(x) \):

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

### Step 5: Evaluate \( f(0) \)
Now, we need to evaluate \( f(0) \):

\[ f(0) = 1 + 0^2 = 1 \]

### Step 6: Check continuity and differentiability at \( x = 0 \)
The function \( f(x) = 1 + x^2 \) is a polynomial function, which is continuous and differentiable everywhere, including at \( x = 0 \).

### Conclusion
Thus, the correct option is that \( f(x) \) is differentiable at \( x = 0 \).

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