The value of `lim_(nto oo)(1^(3)+2^(3)+3^(3)+……..+n^(3))/((n^(2)+1)^(2))`

Options

A. `1/4`
B. `1/2`
C. `1/(2sqrt(2))`
D. None of these

Answer

To solve the limit problem \( \lim_{n \to \infty} \frac{1^3 + 2^3 + 3^3 + \ldots + n^3}{(n^2 + 1)^2} \), we will follow these steps:

### Step 1: Use the formula for the sum of cubes
The sum of the first \( n \) cubes can be expressed using the formula:
\[ 1^3 + 2^3 + 3^3 + \ldots + n^3 = \left( \frac{n(n+1)}{2} \right)^2 \]
Thus, we can rewrite the limit as:
\[ \lim_{n \to \infty} \frac{\left( \frac{n(n+1)}{2} \right)^2}{(n^2 + 1)^2} \]

### Step 2: Simplify the expression
Now, substituting the sum of cubes into the limit gives:
\[ \lim_{n \to \infty} \frac{\left( \frac{n(n+1)}{2} \right)^2}{(n^2 + 1)^2} = \lim_{n \to \infty} \frac{n^2(n+1)^2}{4(n^2 + 1)^2} \]

### Step 3: Expand the numerator and denominator
Expanding the numerator:
\[ n^2(n+1)^2 = n^2(n^2 + 2n + 1) = n^4 + 2n^3 + n^2 \]
And the denominator:
\[ (n^2 + 1)^2 = n^4 + 2n^2 + 1 \]
Thus, we have:
\[ \lim_{n \to \infty} \frac{n^4 + 2n^3 + n^2}{4(n^4 + 2n^2 + 1)} \]

### Step 4: Divide by the highest power of \( n \)
To evaluate the limit as \( n \to \infty \), we divide both the numerator and the denominator by \( n^4 \):
\[ \lim_{n \to \infty} \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{4(1 + \frac{2}{n^2} + \frac{1}{n^4})} \]

### Step 5: Evaluate the limit
As \( n \to \infty \), the terms \( \frac{2}{n} \), \( \frac{1}{n^2} \), \( \frac{2}{n^2} \), and \( \frac{1}{n^4} \) all approach 0. Therefore, we have:
\[ \lim_{n \to \infty} \frac{1 + 0 + 0}{4(1 + 0 + 0)} = \frac{1}{4} \]

### Final Answer
Thus, the value of the limit is:
\[ \frac{1}{4} \]