`lim_(x->oo)((1^(1/x) +2^(1/x) +3^(1/x) +...+n^(1/x))/n)^(nx)` is equal to

Options

A. `|___(n)`
B. `n`
C. `|___(n-1)`
D. 0

Answer

To solve the limit

\[ \lim_{x \to \infty} \left( \frac{1^{1/x} + 2^{1/x} + 3^{1/x} + \ldots + n^{1/x}}{n} \right)^{nx}, \]

we will break it down step by step.

### Step 1: Analyze the terms inside the limit
As \( x \to \infty \), each term \( k^{1/x} \) (where \( k = 1, 2, \ldots, n \)) approaches 1 because:

\[ k^{1/x} = e^{\frac{\log k}{x}} \to e^0 = 1. \]

Thus, we can write:

\[ 1^{1/x} + 2^{1/x} + 3^{1/x} + \ldots + n^{1/x} \to 1 + 1 + 1 + \ldots + 1 = n. \]

### Step 2: Substitute into the limit
Substituting this into our limit expression gives:

\[ \frac{1^{1/x} + 2^{1/x} + 3^{1/x} + \ldots + n^{1/x}}{n} \to \frac{n}{n} = 1. \]

### Step 3: Rewrite the limit
Now we have:

\[ \lim_{x \to \infty} \left( 1 \right)^{nx} = 1^{nx} = 1. \]

### Step 4: Apply the indeterminate form
However, we need to be careful because we have an indeterminate form of \( 1^{\infty} \). To resolve this, we can use the fact that:

\[ \lim_{x \to \infty} (f(x))^{g(x)} = e^{\lim_{x \to \infty} g(x) \cdot (f(x) - 1)}, \]

where \( f(x) \to 1 \) and \( g(x) \to \infty \).

Here, let:

\[ f(x) = \frac{1^{1/x} + 2^{1/x} + 3^{1/x} + \ldots + n^{1/x}}{n} \quad \text{and} \quad g(x) = nx. \]

### Step 5: Calculate \( f(x) - 1 \)
We can express \( f(x) - 1 \):

\[ f(x) - 1 = \frac{1^{1/x} + 2^{1/x} + \ldots + n^{1/x} - n}{n}. \]

### Step 6: Analyze the limit of \( g(x)(f(x) - 1) \)
Now we need to compute:

\[ \lim_{x \to \infty} nx \cdot \left( \frac{1^{1/x} + 2^{1/x} + \ldots + n^{1/x} - n}{n} \right). \]

Using the approximation \( k^{1/x} \approx 1 + \frac{\log k}{x} \) for large \( x \):

\[ 1^{1/x} + 2^{1/x} + \ldots + n^{1/x} \approx n + \frac{\log 1 + \log 2 + \ldots + \log n}{x} = n + \frac{\log(n!)}{x}. \]

Thus,

\[ f(x) - 1 \approx \frac{\frac{\log(n!)}{x}}{n} = \frac{\log(n!)}{nx}. \]

### Step 7: Substitute back into the limit
Now substituting back gives:

\[ \lim_{x \to \infty} nx \cdot \frac{\log(n!)}{nx} = \log(n!). \]

### Step 8: Final result
Therefore, we have:

\[ \lim_{x \to \infty} (f(x))^{g(x)} = e^{\log(n!)} = n!. \]

Thus, the final answer is:

\[ \boxed{n!}. \]