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

Options

A. has no limit
B. is discontinuous
C. is continuous but not differentiable
D. is differentiable

Answer

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

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

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

### Step 1: Identify the first term and the common ratio
The first term of the series is \( a = x \) and the common ratio \( r \) can be identified as \( \frac{1}{1+x} \).

### Step 2: Write the sum of the infinite 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, we have:

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

### Step 3: Write the function \( f(x) \)
Thus, we can express \( f(x) \) as:

\[ f(x) = 1 + x \]

### Step 4: Evaluate \( f(x) \) at \( x = 0 \)
Now, we need to evaluate \( f(x) \) at \( x = 0 \):

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

### Step 5: Analyze the continuity of \( f(x) \)
The function \( f(x) = 1 + x \) is a linear function, which is continuous everywhere, including at \( x = 0 \).

### Conclusion
Thus, at \( x = 0 \), the function \( f(x) \) is defined and equals 1.

### Final Answer
The value of \( f(0) \) is \( 1 \).

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