`lim_(ntooo) (n(2n+1)^(2))/((n+2)(n^(2)+3n-1))" is equal to "`

Options

A. 1
B. 0
C. 2
D. none of these

Explanation

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

### Step 1: Rewrite the limit expression
We start with the limit:
\[ L = \lim_{n \to \infty} \frac{n(2n+1)^2}{(n+2)(n^2 + 3n - 1)} \]

### Step 2: Expand the numerator
The numerator can be expanded:
\[ (2n + 1)^2 = 4n^2 + 4n + 1 \]
Thus, the numerator becomes:
\[ n(2n + 1)^2 = n(4n^2 + 4n + 1) = 4n^3 + 4n^2 + n \]

### Step 3: Expand the denominator
Now we expand the denominator:
\[ (n + 2)(n^2 + 3n - 1) = n(n^2 + 3n - 1) + 2(n^2 + 3n - 1) = n^3 + 3n^2 - n + 2n^2 + 6n - 2 \]
Combining like terms, we get:
\[ n^3 + 5n^2 + 5n - 2 \]

### Step 4: Substitute back into the limit
Now we substitute the expanded forms back into the limit:
\[ L = \lim_{n \to \infty} \frac{4n^3 + 4n^2 + n}{n^3 + 5n^2 + 5n - 2} \]

### Step 5: Divide numerator and denominator by \( n^3 \)
Next, we divide every term in the numerator and denominator by \( n^3 \):
\[ L = \lim_{n \to \infty} \frac{4 + \frac{4}{n} + \frac{1}{n^2}}{1 + \frac{5}{n} + \frac{5}{n^2} - \frac{2}{n^3}} \]

### Step 6: Evaluate the limit as \( n \to \infty \)
As \( n \) approaches infinity, the terms \( \frac{4}{n} \), \( \frac{1}{n^2} \), \( \frac{5}{n} \), \( \frac{5}{n^2} \), and \( \frac{2}{n^3} \) all approach 0:
\[ L = \frac{4 + 0 + 0}{1 + 0 + 0 - 0} = \frac{4}{1} = 4 \]

### Conclusion
Thus, the limit is:
\[ \boxed{4} \]