The value of `lim_(nto oo){3sqrt(n^2-n^3)+n}`, is

Options

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

Explanation

To solve the limit \( \lim_{n \to \infty} \left( 3\sqrt{n^2 - n^3} + n \right) \), we can follow these steps:

### Step 1: Rewrite the expression
Start by rewriting the expression inside the limit:
\[ \lim_{n \to \infty} \left( 3\sqrt{n^2 - n^3} + n \right) \]

### Step 2: Factor out \( n^3 \) from the square root
Notice that we can factor \( n^3 \) out of the square root:
\[ \sqrt{n^2 - n^3} = \sqrt{n^3 \left( \frac{1}{n} - 1 \right)} = n\sqrt{\frac{1}{n} - 1} \]
Thus, we can rewrite the limit as:
\[ \lim_{n \to \infty} \left( 3n\sqrt{\frac{1}{n} - 1} + n \right) \]

### Step 3: Factor out \( n \)
Now, factor \( n \) out of the entire expression:
\[ \lim_{n \to \infty} n \left( 3\sqrt{\frac{1}{n} - 1} + 1 \right) \]

### Step 4: Analyze the limit of the square root term
As \( n \to \infty \), \( \frac{1}{n} \) approaches \( 0 \):
\[ \sqrt{\frac{1}{n} - 1} \to \sqrt{0 - 1} = \sqrt{-1} \text{ (not defined in real numbers)} \]
However, we can analyze the limit more carefully:
\[ \sqrt{\frac{1}{n} - 1} = \sqrt{-1 + \frac{1}{n}} \to \sqrt{-1} \text{ (which is } i \text{ in complex numbers)} \]
But for our limit, we need to consider the behavior as \( n \to \infty \):
\[ \sqrt{\frac{1}{n} - 1} \approx \sqrt{-1} \text{ (dominates)} \]

### Step 5: Substitute back into the limit
Now substitute back into the limit:
\[ \lim_{n \to \infty} n \left( 3\sqrt{-1} + 1 \right) = \lim_{n \to \infty} n \left( 3i + 1 \right) \]

### Step 6: Conclude the limit
As \( n \to \infty \), the term \( n(3i + 1) \) diverges unless \( 3i + 1 = 0 \), which it does not. Thus, the limit diverges.

### Final Answer
The limit does not converge to a finite value; it diverges.