`lim_(n to oo)((1)/(1+n^(3))+(4)/(8+n^(3))+....+(r^(2))/(r^(3)+n^(3))+....+(1)/(2*n^3)`

Options

A. `(1)/(2)log 2`
B. `(1)/(3)log 2`
C. `(1)/(4) log 2`
D. `log 2`

Answer

To solve the limit

\[ \lim_{n \to \infty} \left( \frac{1}{1+n^3} + \frac{4}{8+n^3} + \ldots + \frac{r^2}{r^3+n^3} + \ldots + \frac{1}{2n^3} \right), \]

we can express this as a summation:

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

### Step 1: Rewrite the Summation
We can factor out \(n^3\) from the denominator:

\[ \frac{r^2}{r^3 + n^3} = \frac{r^2/n^3}{r^3/n^3 + 1} = \frac{\left(\frac{r}{n}\right)^2}{\left(\frac{r}{n}\right)^3 + 1}. \]

### Step 2: Change of Variables
Let \(x = \frac{r}{n}\). As \(n \to \infty\), \(r\) goes from \(1\) to \(n\), which means \(x\) goes from \(\frac{1}{n}\) to \(1\). The increment in \(r\) corresponds to an increment in \(x\) of \(\frac{1}{n}\). Thus, we can rewrite the sum as:

\[ \sum_{r=1}^{n} \frac{r^2/n^3}{r^3/n^3 + 1} \cdot \frac{1}{n} \approx \int_{0}^{1} \frac{x^2}{x^3 + 1} \, dx. \]

### Step 3: Set Up the Integral
Now we need to evaluate the integral:

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

### Step 4: Use Substitution
Let \(t = x^3 + 1\). Then, \(dt = 3x^2 \, dx\) or \(dx = \frac{dt}{3x^2}\). When \(x = 0\), \(t = 1\) and when \(x = 1\), \(t = 2\). Thus, we can rewrite the integral:

\[ \int_{1}^{2} \frac{x^2}{t} \cdot \frac{dt}{3x^2} = \frac{1}{3} \int_{1}^{2} \frac{1}{t} \, dt. \]

### Step 5: Evaluate the Integral
The integral \(\int \frac{1}{t} \, dt\) is \(\ln(t)\). Therefore, we have:

\[ \frac{1}{3} \left[ \ln(t) \right]_{1}^{2} = \frac{1}{3} (\ln(2) - \ln(1)) = \frac{1}{3} \ln(2). \]

### Final Answer
Thus, the limit evaluates to:

\[ \lim_{n \to \infty} \left( \frac{1}{1+n^3} + \frac{4}{8+n^3} + \ldots + \frac{1}{2n^3} \right) = \frac{1}{3} \ln(2). \]

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