If `f(x) =x^(2) + x^(2)/(1+x^(2)) + x^(2)/(1+x^(2))^(2) + …… + x^(2)/(1+x^(2))^(n)` + ….. Then at x =0

Options

A. f (x) has no limit
B. f(x) is discontinuous
C. f(x) is continuous but not differentiable
D. f(x) is differentiable

Answer

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

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

### Step 1: Rewrite the function
We can factor out \( x^2 \) from each term in the series:

\[ f(x) = x^2 \left( 1 + \frac{1}{1+x^2} + \frac{1}{(1+x^2)^2} + \cdots \right) \]

### Step 2: Identify the series
The series inside the parentheses is a geometric series with the first term \( a = 1 \) and the common ratio \( r = \frac{1}{1+x^2} \).

### Step 3: Sum the geometric series
The sum of an infinite geometric series is given by:

\[ S = \frac{a}{1 - r} \]

Thus, we have:

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

### Step 4: Substitute back into \( f(x) \)
Now substituting back into \( f(x) \):

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

### Step 5: Evaluate \( f(0) \)
Now we need to evaluate \( f(x) \) at \( x = 0 \):

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

### Step 6: Check continuity at \( x = 0 \)
To check if \( f(x) \) is continuous at \( x = 0 \), we need to find the left-hand limit and the right-hand limit as \( x \) approaches \( 0 \).

- **Left-hand limit**:
\[ \lim_{h \to 0^-} f(0 - h) = \lim_{h \to 0^-} (1 + h^2) = 1 \]

- **Right-hand limit**:
\[ \lim_{h \to 0^+} f(0 + h) = \lim_{h \to 0^+} (1 + h^2) = 1 \]

Since both limits equal \( f(0) = 1 \), \( f(x) \) is continuous at \( x = 0 \).

### Step 7: Check differentiability at \( x = 0 \)
To check if \( f(x) \) is differentiable at \( x = 0 \), we compute the left-hand derivative and right-hand derivative.

- **Right-hand derivative**:
\[ f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{(1 + h^2) - 1}{h} = \lim_{h \to 0} \frac{h^2}{h} = \lim_{h \to 0} h = 0 \]

- **Left-hand derivative**:
\[ f'(0) = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{-h} = \lim_{h \to 0} \frac{(1 + h^2) - 1}{-h} = \lim_{h \to 0} \frac{h^2}{-h} = \lim_{h \to 0} -h = 0 \]

Since both derivatives are equal, \( f(x) \) is differentiable at \( x = 0 \).

### Conclusion
Thus, we conclude that \( f(x) \) is continuous and differentiable at \( x = 0 \).