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 problem, we will break it down step by step.

### Step 1: Rewrite the Limit Expression
We start with the limit expression:
\[ \lim_{n \to \infty} \left( \frac{1^2}{1^3 + n^3} + \frac{2^2}{2^3 + n^3} + \ldots + \frac{n^2}{n^3 + n^3} \right) \]
This can be expressed using summation notation:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{r^2}{r^3 + n^3} \]

### Step 2: Simplify the General Term
We can simplify the term inside the summation:
\[ \frac{r^2}{r^3 + n^3} = \frac{r^2/n^3}{r^3/n^3 + 1} = \frac{\left(\frac{r}{n}\right)^2}{\left(\frac{r}{n}\right)^3 + 1} \]
Let \( x = \frac{r}{n} \). As \( n \to \infty \), \( r \) varies from \( 1 \) to \( n \), thus \( x \) varies from \( \frac{1}{n} \) to \( 1 \).

### Step 3: Change the Summation to an Integral
The sum can be approximated by an integral:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{x^2}{x^3 + 1} \cdot \frac{1}{n} \quad \text{(where } dx = \frac{1}{n} \text{)} \]
This leads us to:
\[ \int_{0}^{1} \frac{x^2}{x^3 + 1} \, dx \]

### Step 4: Solve the Integral
Now we need to evaluate the integral:
\[ \int_{0}^{1} \frac{x^2}{x^3 + 1} \, dx \]
We can use the substitution \( u = x^3 + 1 \), then \( du = 3x^2 \, dx \) or \( dx = \frac{du}{3x^2} \).

When \( x = 0 \), \( u = 1 \) and when \( x = 1 \), \( u = 2 \).

Thus, the integral becomes:
\[ \int_{1}^{2} \frac{x^2}{u} \cdot \frac{du}{3x^2} = \frac{1}{3} \int_{1}^{2} \frac{1}{u} \, du \]
This evaluates to:
\[ \frac{1}{3} \left[ \ln u \right]_{1}^{2} = \frac{1}{3} (\ln 2 - \ln 1) = \frac{1}{3} \ln 2 \]

### Step 5: Final Result
Thus, the limit evaluates to:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{r^2}{r^3 + n^3} = \frac{1}{3} \ln 2 \]

### Conclusion
The value of the limit is:
\[ \frac{1}{3} \ln 2 \]