The value of `lim_(n to oo)((n!)/(n^(n)))^((2n^(4)+1)/(5n^(5)+1))` is equal

Options

A. e
B. `(2)/(e)`
C. `((1)/(e))^((2)/(5))`
D. None of these

Answer

To solve the limit \( \lim_{n \to \infty} \left( \frac{n!}{n^n} \right)^{\frac{2n^4 + 1}{5n^5 + 1}} \), we will follow these steps:

### Step 1: Rewrite the limit
We start with the expression:
\[ L = \lim_{n \to \infty} \left( \frac{n!}{n^n} \right)^{\frac{2n^4 + 1}{5n^5 + 1}} \]

### Step 2: Use Stirling's approximation
Using Stirling's approximation, we have:
\[ n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \]
Thus, we can rewrite \( \frac{n!}{n^n} \):
\[ \frac{n!}{n^n} \sim \frac{\sqrt{2 \pi n} \left( \frac{n}{e} \right)^n}{n^n} = \frac{\sqrt{2 \pi n}}{e^n} \]

### Step 3: Substitute into the limit
Now substituting this back into the limit:
\[ L = \lim_{n \to \infty} \left( \frac{\sqrt{2 \pi n}}{e^n} \right)^{\frac{2n^4 + 1}{5n^5 + 1}} \]

### Step 4: Simplify the exponent
We can simplify the exponent:
\[ \frac{2n^4 + 1}{5n^5 + 1} \sim \frac{2n^4}{5n^5} = \frac{2}{5n} \]
Thus, we have:
\[ L = \lim_{n \to \infty} \left( \frac{\sqrt{2 \pi n}}{e^n} \right)^{\frac{2}{5n}} \]

### Step 5: Break it down
Now we can break this limit into two parts:
\[ L = \lim_{n \to \infty} \left( \sqrt{2 \pi n} \right)^{\frac{2}{5n}} \cdot \left( e^{-n} \right)^{\frac{2}{5n}} \]

### Step 6: Evaluate each part
1. For the first part:
\[ \left( \sqrt{2 \pi n} \right)^{\frac{2}{5n}} = (2 \pi n)^{\frac{1}{5n}} = e^{\frac{1}{5n} \log(2 \pi n)} \to e^0 = 1 \quad \text{as } n \to \infty \]

2. For the second part:
\[ \left( e^{-n} \right)^{\frac{2}{5n}} = e^{-\frac{2}{5}} \quad \text{as } n \to \infty \]

### Step 7: Combine the results
Combining both parts, we get:
\[ L = 1 \cdot e^{-\frac{2}{5}} = e^{-\frac{2}{5}} \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{\frac{1}{e^{\frac{2}{5}}}} \]