For `x in R -{-1/n, n in N}`, define `f(x) = lim_(n to infty)(x/(x+1) + x/((x+1)(2x+1)) + x/((2x+1)(3x+1)))` + ….. + upto n terms
then range of f contains exactly……….. Element(s).

Answer

To solve the problem, we need to evaluate the function \( f(x) \) defined as:

\[ f(x) = \lim_{n \to \infty} \left( \frac{x}{x+1} + \frac{x}{(x+1)(2x+1)} + \frac{x}{(2x+1)(3x+1)} + \ldots + \text{ up to } n \text{ terms} \right) \]

### Step 1: Understand the Domain
The function is defined for \( x \in \mathbb{R} \setminus \{-\frac{1}{n} : n \in \mathbb{N}\} \). This means that \( x \) can take any real value except for the values of the form \( -\frac{1}{n} \) where \( n \) is a natural number.

**Hint:** Identify the values of \( x \) that are excluded from the domain.

### Step 2: Write the General Term
The general term in the series can be expressed as:

\[ \frac{x}{(kx + 1)((k+1)x + 1)} \text{ for } k = 0, 1, 2, \ldots, n-1 \]

### Step 3: Simplifying the Series
We can rewrite the sum as:

\[ \sum_{k=0}^{n-1} \frac{x}{(kx + 1)((k+1)x + 1)} \]

This can be simplified using partial fractions. We can express:

\[ \frac{x}{(kx + 1)((k+1)x + 1)} = \frac{A}{kx + 1} + \frac{B}{(k+1)x + 1} \]

### Step 4: Finding the Limit
As \( n \to \infty \), we observe that the series converges. The limit can be evaluated by noticing that the terms will cancel out in a telescoping manner.

The limit simplifies to:

\[ \lim_{n \to \infty} \left( 1 - \frac{1}{nx + 1} \right) \]

### Step 5: Evaluate the Limit
As \( n \to \infty \), the term \( \frac{1}{nx + 1} \) approaches 0, thus:

\[ f(x) = 1 \]

### Conclusion
Since \( f(x) = 1 \) for all \( x \) in the domain, the range of \( f \) contains exactly **1 element**.

**Final Answer:** The range of \( f \) contains exactly **1 element**.