If `f(x)=(x)/(1+x)+(x)/((x+1)(2x+1))+(x)/((2x+1)(3x+1))+…"to "infty`,then at x = 0,f(x)
(a) has no limit
(b) is discontinuous
(c) is continuous but not differentiable
(d) is differentiable

Options

A. 1
B. 2
C. 3
D. 4

Answer

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

\[ f(x) = \frac{x}{1+x} + \frac{x}{(x+1)(2x+1)} + \frac{x}{(2x+1)(3x+1)} + \ldots \]

We will evaluate the behavior of \( f(x) \) as \( x \) approaches 0.

### Step 1: Rewrite the function

The function can be expressed as a series:

\[ f(x) = \sum_{n=1}^{\infty} \frac{x}{(nx + 1)((n+1)x + 1)} \]

### Step 2: Evaluate \( f(0) \)

To find \( f(0) \), we substitute \( x = 0 \):

\[ f(0) = \sum_{n=1}^{\infty} \frac{0}{(n \cdot 0 + 1)((n+1) \cdot 0 + 1)} = \sum_{n=1}^{\infty} 0 = 0 \]

### Step 3: Check continuity at \( x = 0 \)

To check the continuity of \( f(x) \) at \( x = 0 \), we need to find the limit of \( f(x) \) as \( x \) approaches 0:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{x}{1+x} + \frac{x}{(x+1)(2x+1)} + \frac{x}{(2x+1)(3x+1)} + \ldots \right) \]

Each term in the series approaches 0 as \( x \) approaches 0. Therefore, the limit is:

\[ \lim_{x \to 0} f(x) = 0 \]

Since \( f(0) = 0 \) and \( \lim_{x \to 0} f(x) = 0 \), we conclude that \( f(x) \) is continuous at \( x = 0 \).

### Step 4: Check differentiability at \( x = 0 \)

To check if \( f(x) \) is differentiable at \( x = 0 \), we need to find the derivative \( f'(x) \):

Using the series representation, we can differentiate term by term:

\[ f'(x) = \sum_{n=1}^{\infty} \frac{d}{dx} \left( \frac{x}{(nx + 1)((n+1)x + 1)} \right) \]

Calculating the derivative for each term will show that it is well-defined and continuous around \( x = 0 \). Thus, \( f'(0) \) exists, indicating that \( f(x) \) is differentiable at \( x = 0 \).

### Conclusion

Since \( f(x) \) is continuous and differentiable at \( x = 0 \), the correct answer is:

**(d) is differentiable.**

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