The value of `lim_(nrarroo)([x]+[2^(2)x]+[3^(2)x]+…+[n^(2)x])/(1^(2)+2^(2)+3^(2)+….+n^(2))` is equal to (where `[x]` represents the greatest integer part of x)

Options

A. x
B. 2x
C. `(x)/(2)`
D. `(x)/(6)`

Answer

To solve the limit problem given by

\[ \lim_{n \to \infty} \frac{[x] + [2^2 x] + [3^2 x] + \ldots + [n^2 x]}{1^2 + 2^2 + 3^2 + \ldots + n^2} \]

where \([x]\) represents the greatest integer part of \(x\), we can follow these steps:

### Step 1: Understand the Components

The numerator consists of the sum of the greatest integer parts of \(k^2 x\) for \(k = 1\) to \(n\). The denominator is the sum of squares of the first \(n\) natural numbers.

### Step 2: Analyze the Denominator

The sum of squares of the first \(n\) natural numbers can be expressed as:

\[ 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 3: Analyze the Numerator

For the numerator, we can express it as:

\[ \sum_{k=1}^{n} [k^2 x] \]

Using the property of the greatest integer function, we know that:

\[ [k^2 x] = k^2 x - \{k^2 x\} \]

where \(\{k^2 x\}\) is the fractional part of \(k^2 x\). Therefore, we can write:

\[ \sum_{k=1}^{n} [k^2 x] = \sum_{k=1}^{n} (k^2 x - \{k^2 x\}) = x \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} \{k^2 x\} \]

### Step 4: Substitute the Denominator

Now substituting the expression for the sum of squares into our limit, we have:

\[ \lim_{n \to \infty} \frac{x \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} \{k^2 x\}}{\frac{n(n + 1)(2n + 1)}{6}} \]

### Step 5: Simplify the Limit

As \(n\) approaches infinity, the term \(\sum_{k=1}^{n} k^2\) behaves like \(\frac{n^3}{3}\) and thus:

\[ \sum_{k=1}^{n} k^2 \sim \frac{n^3}{3} \]

This leads to:

\[ \lim_{n \to \infty} \frac{x \cdot \frac{n^3}{3} - \sum_{k=1}^{n} \{k^2 x\}}{\frac{n(n + 1)(2n + 1)}{6}} \approx \lim_{n \to \infty} \frac{2x n^3}{n^3} = 2x \]

### Step 6: Evaluate the Limit

Since the fractional part \(\sum_{k=1}^{n} \{k^2 x\}\) grows slower than \(n^3\), we can ignore it in the limit, leading to:

\[ \lim_{n \to \infty} \frac{[x] + [2^2 x] + [3^2 x] + \ldots + [n^2 x]}{1^2 + 2^2 + 3^2 + \ldots + n^2} = x \]

### Conclusion

Thus, the final result is:

\[ \lim_{n \to \infty} \frac{[x] + [2^2 x] + [3^2 x] + \ldots + [n^2 x]}{1^2 + 2^2 + 3^2 + \ldots + n^2} = x \]