The value of `lim _(xto oo) ((n !)/(n ^(n)))^((3n^(3)+4)/(4n ^(4)-1)), n inN` is equal to:

Options

A. `((1)/(e ))^(3//4)`
B. `e ^(3//4)`
C. `e ^(-1)`
D. 0

Answer

To solve the limit problem, we need to evaluate:

\[ \lim_{n \to \infty} \left( \frac{n!}{n^n} \right)^{\frac{3n^3 + 4}{4n^4 - 1}} \]

### Step 1: Rewrite the limit expression
Let \( y = \lim_{n \to \infty} \left( \frac{n!}{n^n} \right)^{\frac{3n^3 + 4}{4n^4 - 1}} \).

### Step 2: Take the natural logarithm
Taking the natural logarithm of both sides gives us:

\[ \ln y = \lim_{n \to \infty} \frac{3n^3 + 4}{4n^4 - 1} \ln \left( \frac{n!}{n^n} \right) \]

### Step 3: Simplify the logarithm
Using Stirling's approximation, we know that \( n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \). Therefore,

\[ \ln(n!) \sim \frac{1}{2} \ln(2 \pi n) + n - n \ln n \]

So,

\[ \ln \left( \frac{n!}{n^n} \right) = \ln(n!) - n \ln n \sim \frac{1}{2} \ln(2 \pi n) + n - n \ln n - n \ln n \]

This simplifies to:

\[ \ln \left( \frac{n!}{n^n} \right) \sim \frac{1}{2} \ln(2 \pi n) - 2n \ln n \]

### Step 4: Substitute back into the limit
Now substituting this back into our expression for \( \ln y \):

\[ \ln y = \lim_{n \to \infty} \frac{3n^3 + 4}{4n^4 - 1} \left( \frac{1}{2} \ln(2 \pi n) - 2n \ln n \right) \]

### Step 5: Analyze the limit
As \( n \to \infty \), the dominant term in the logarithm will be \( -2n \ln n \). Thus, we can focus on that term:

\[ \ln y \sim \lim_{n \to \infty} \frac{3n^3 + 4}{4n^4 - 1} (-2n \ln n) \]

### Step 6: Simplify the fraction
The limit of the fraction as \( n \to \infty \) is:

\[ \lim_{n \to \infty} \frac{3n^3 + 4}{4n^4 - 1} = \lim_{n \to \infty} \frac{3/n + 4/n^3}{4 - 1/n^4} = 0 \]

### Step 7: Conclude the limit
Thus,

\[ \ln y = 0 \implies y = e^0 = 1 \]

### Final Answer
The value of the limit is:

\[ \lim_{n \to \infty} \left( \frac{n!}{n^n} \right)^{\frac{3n^3 + 4}{4n^4 - 1}} = 1 \]