The value of `lim_(n to oo)((1)/(1^(3)+n^(3))+(2^(2))/(2^(3)+n^(3))+..........+(n^(2))/(n^(3)+n^(3)))` is :

Options

A. `(1)/(3)`
B. `(1)/(3)log(2)`
C. `(1)/(2)log(3)`
D. `(1)/(3)log(3)`

Answer

To solve the limit

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

we can express the limit in a more manageable form.

### Step 1: Rewrite the expression
The expression can be rewritten as:

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

### Step 2: Factor out \(n^3\) from the denominator
We can factor \(n^3\) from the denominator:

\[ = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k^2}{n^3 \left(\frac{k^3}{n^3} + 1\right)} = \lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^{n} \frac{k^2}{\left(\frac{k}{n}\right)^3 + 1}. \]

### Step 3: Change the variable
Let \(x = \frac{k}{n}\), then \(k = nx\) and \(dk = n \, dx\). The sum can be approximated by an integral as \(n\) approaches infinity:

\[ \sum_{k=1}^{n} \frac{k^2}{\left(\frac{k}{n}\right)^3 + 1} \approx n \int_{0}^{1} \frac{(nx)^2}{x^3 + 1} \, dx. \]

### Step 4: Substitute back into the limit
Substituting this back into our limit gives:

\[ \lim_{n \to \infty} \frac{1}{n^3} n \int_{0}^{1} \frac{(nx)^2}{x^3 + 1} \, dx = \lim_{n \to \infty} \frac{n^2}{n^3} \int_{0}^{1} \frac{(nx)^2}{x^3 + 1} \, dx = \lim_{n \to \infty} \frac{1}{n} \int_{0}^{1} \frac{n^2 x^2}{x^3 + 1} \, dx. \]

### Step 5: Evaluate the integral
Now we can evaluate the integral:

\[ \int_{0}^{1} \frac{n^2 x^2}{x^3 + 1} \, dx. \]

As \(n\) approaches infinity, we can factor out \(n^2\):

\[ = n^2 \int_{0}^{1} \frac{x^2}{x^3 + 1} \, dx. \]

### Step 6: Final limit
Thus, the limit simplifies to:

\[ \lim_{n \to \infty} \frac{1}{n} \cdot n^2 \cdot \int_{0}^{1} \frac{x^2}{x^3 + 1} \, dx = \lim_{n \to \infty} n \cdot \int_{0}^{1} \frac{x^2}{x^3 + 1} \, dx. \]

Since \(\int_{0}^{1} \frac{x^2}{x^3 + 1} \, dx\) is a constant, the limit approaches \(0\).

### Conclusion
Thus, the value of the limit is:

\[ \boxed{0}. \]