If `[x]` denotes the greatest integer less than or equal to x, then evaluate `lim_(ntooo) (1)/(n^(2))([1.x]+[2.x]+[3.x]+...+[n.x]).`

Explanation

To evaluate the limit

\[ \lim_{n \to \infty} \frac{1}{n^2} \left( [1 \cdot x] + [2 \cdot x] + [3 \cdot x] + \ldots + [n \cdot x] \right), \]

where \([x]\) denotes the greatest integer less than or equal to \(x\), we can follow these steps:

### Step 1: Rewrite the Sum
We can express the sum in terms of a summation:

\[ \sum_{r=1}^{n} [r \cdot x]. \]

Using the property of the greatest integer function, we have:

\[ [r \cdot x] = r \cdot x - \{r \cdot x\}, \]

where \(\{r \cdot x\}\) is the fractional part of \(r \cdot x\). Thus, we can rewrite the sum as:

\[ \sum_{r=1}^{n} [r \cdot x] = \sum_{r=1}^{n} (r \cdot x - \{r \cdot x\}) = \sum_{r=1}^{n} r \cdot x - \sum_{r=1}^{n} \{r \cdot x\}. \]

### Step 2: Substitute into the Limit
Now substituting this back into the limit gives us:

\[ \lim_{n \to \infty} \frac{1}{n^2} \left( \sum_{r=1}^{n} r \cdot x - \sum_{r=1}^{n} \{r \cdot x\} \right). \]

### Step 3: Evaluate Each Sum
The first sum can be simplified using the formula for the sum of the first \(n\) natural numbers:

\[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2}. \]

Thus,

\[ \sum_{r=1}^{n} r \cdot x = x \cdot \frac{n(n+1)}{2}. \]

Now substituting this back into the limit:

\[ \lim_{n \to \infty} \frac{1}{n^2} \left( x \cdot \frac{n(n+1)}{2} - \sum_{r=1}^{n} \{r \cdot x\} \right). \]

### Step 4: Analyze the Fractional Part Sum
The sum \(\sum_{r=1}^{n} \{r \cdot x\}\) consists of fractional parts, which are bounded between \(0\) and \(1\). Therefore, as \(n\) approaches infinity, this sum grows linearly with \(n\), but is significantly smaller than \(n^2\):

\[ \sum_{r=1}^{n} \{r \cdot x\} \leq n. \]

### Step 5: Simplify the Limit
Now we can simplify the limit:

\[ \lim_{n \to \infty} \left( \frac{x \cdot \frac{n(n+1)}{2}}{n^2} - \frac{\sum_{r=1}^{n} \{r \cdot x\}}{n^2} \right). \]

The first term simplifies to:

\[ \frac{x(n^2 + n)}{2n^2} = \frac{x}{2} + \frac{x}{2n}. \]

As \(n\) approaches infinity, the second term \(\frac{\sum_{r=1}^{n} \{r \cdot x\}}{n^2}\) tends to \(0\).

### Final Result
Thus, we find:

\[ \lim_{n \to \infty} \left( \frac{x}{2} + \frac{x}{2n} - 0 \right) = \frac{x}{2}. \]

Therefore, the final answer is:

\[ \boxed{\frac{x}{2}}. \]