The value of `lim_(xtooo){lim_(ntooo)([1^(2)(sinx)^(x)]+[2^(2)(sinx)^(x)]+……….+[n^(2)(sinx)^(x)])/(n^(3))}` is (wehre [.] denotes the greatest integer function)

Options

A. A. `x/3+(sinx)/3`
B. B. `x/3+(sinx)^(x)`
C. C. `1/3`
D. D. 0

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{x \to 0^+} \lim_{n \to \infty} \frac{[1^2 \sin^x x] + [2^2 \sin^x x] + \ldots + [n^2 \sin^x x]}{n^3} \]

### Step 1: Rewrite the expression

We can express the sum in the numerator as:

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

Thus, we can rewrite the limit as:

\[ \lim_{x \to 0^+} \lim_{n \to \infty} \frac{\sin^x x \sum_{k=1}^{n} k^2}{n^3} \]

### Step 2: Use the formula for the sum of squares

The formula for the sum of the squares of the first \( n \) natural numbers is:

\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]

Substituting this into our expression gives:

\[ \lim_{x \to 0^+} \lim_{n \to \infty} \frac{\sin^x x \cdot \frac{n(n+1)(2n+1)}{6}}{n^3} \]

### Step 3: Simplify the expression

Now, simplifying the fraction:

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

As \( n \to \infty \), the terms \( \frac{3}{n} \) and \( \frac{1}{n^2} \) approach 0, so we have:

\[ \lim_{n \to \infty} \frac{2 + \frac{3}{n} + \frac{1}{n^2}}{6} = \frac{2}{6} = \frac{1}{3} \]

### Step 4: Substitute back into the limit

Now, substituting this back into our limit gives:

\[ \lim_{x \to 0^+} \sin^x x \cdot \frac{1}{3} \]

### Step 5: Evaluate \( \sin^x x \)

Next, we need to evaluate \( \sin^x x \) as \( x \to 0^+ \). We know that:

\[ \sin x \approx x \quad \text{as } x \to 0 \]

Thus:

\[ \sin^x x \approx x^x \]

And we can express \( x^x \) as:

\[ x^x = e^{x \ln x} \]

As \( x \to 0^+ \), \( x \ln x \to 0 \). Therefore:

\[ \lim_{x \to 0^+} x^x = e^0 = 1 \]

### Step 6: Final limit evaluation

Putting everything together, we have:

\[ \lim_{x \to 0^+} \sin^x x \cdot \frac{1}{3} = 1 \cdot \frac{1}{3} = \frac{1}{3} \]

### Step 7: Apply the greatest integer function

Since we need to apply the greatest integer function, we have:

\[ \lfloor \frac{1}{3} \rfloor = 0 \]

Thus, the final answer is:

\[ \boxed{0} \]