`lim_(nrarroo) [(1)/(n)+(n^(2))/((n+1)^(3))+(n^(2))/((n+2)^(3))+...+(1)/(8n)]` is equal to

Options

A. `(3)/(8)`
B. `(1)/(4)`
C. `(1)/(8)`
D. None of these

Explanation

To solve the limit expression given in the question, we will break it down step by step.

### Step 1: Rewrite the Limit Expression
We start with the limit expression:
\[ \lim_{n \to \infty} \left[ \frac{1}{n} + \frac{n^2}{(n+1)^3} + \frac{n^2}{(n+2)^3} + \ldots + \frac{1}{8n} \right] \]
We can express this as:
\[ \lim_{n \to \infty} \sum_{r=1}^{8n} \frac{n^2}{(n+r)^3} \]

### Step 2: Simplify the Terms
Notice that for large \(n\), we can simplify the term \((n+r)^3\):
\[ (n+r)^3 = n^3 \left(1 + \frac{r}{n}\right)^3 \]
Thus, we can rewrite our expression:
\[ \frac{n^2}{(n+r)^3} = \frac{n^2}{n^3 \left(1 + \frac{r}{n}\right)^3} = \frac{1}{n \left(1 + \frac{r}{n}\right)^3} \]

### Step 3: Rewrite the Sum
Now, substituting this back into our sum, we get:
\[ \lim_{n \to \infty} \sum_{r=1}^{8n} \frac{1}{n \left(1 + \frac{r}{n}\right)^3} \]
This can be recognized as a Riemann sum for the function \(f(x) = \frac{1}{(1+x)^3}\) over the interval from \(0\) to \(8\) as \(n\) approaches infinity.

### Step 4: Convert to Integral
As \(n\) approaches infinity, \(\frac{r}{n}\) approaches \(x\), and the sum becomes:
\[ \int_0^8 \frac{1}{(1+x)^3} \, dx \]

### Step 5: Evaluate the Integral
Now we need to evaluate the integral:
\[ \int_0^8 \frac{1}{(1+x)^3} \, dx \]
Using the substitution \(u = 1+x\), we have \(du = dx\) and the limits change from \(1\) to \(9\):
\[ \int_1^9 \frac{1}{u^3} \, du \]
This integral evaluates to:
\[ \left[-\frac{1}{2u^2}\right]_1^9 = -\frac{1}{2(9^2)} + \frac{1}{2(1^2)} = -\frac{1}{162} + \frac{1}{2} \]
Calculating this gives:
\[ \frac{1}{2} - \frac{1}{162} = \frac{81}{162} - \frac{1}{162} = \frac{80}{162} = \frac{40}{81} \]

### Step 6: Final Result
Thus, the limit evaluates to:
\[ \lim_{n \to \infty} \left[ \frac{1}{n} + \frac{n^2}{(n+1)^3} + \frac{n^2}{(n+2)^3} + \ldots + \frac{1}{8n} \right] = \frac{40}{81} \]