`lim_(x->0)((1^x+2^x+3^x+....+n^x)/n)^(1/x)`

Options

A. `f(1^(+))+f(1^(-))=0`
B. `f(1^(+))+f(1^(-))+f(1)=3//2`
C. `f(-1^(+))+f(-1^(-))=-1`
D. `f(1^(+))+f(-1^(-))=0`

Explanation

To solve the limit

\[ \lim_{x \to 0} \left( \frac{1^x + 2^x + 3^x + \ldots + n^x}{n} \right)^{\frac{1}{x}}, \]

we will follow these steps:

### Step 1: Analyze the expression as \( x \to 0 \)

As \( x \) approaches 0, we know that \( a^x \) approaches 1 for any positive number \( a \). Therefore, each term \( k^x \) (where \( k = 1, 2, \ldots, n \)) approaches 1:

\[ 1^x \to 1, \quad 2^x \to 1, \quad \ldots, \quad n^x \to 1. \]

Thus, the sum \( 1^x + 2^x + 3^x + \ldots + n^x \) approaches \( n \) as \( x \to 0 \).

### Step 2: Substitute into the limit

Substituting this result into the limit gives us:

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

So, we have:

\[ \lim_{x \to 0} \left( \frac{1^x + 2^x + 3^x + \ldots + n^x}{n} \right)^{\frac{1}{x}} = \lim_{x \to 0} 1^{\frac{1}{x}}. \]

### Step 3: Recognize the indeterminate form

The expression \( 1^{\frac{1}{x}} \) is an indeterminate form of type \( 1^\infty \). To resolve this, we can use the exponential limit transformation:

\[ \lim_{x \to 0} a^{b} = e^{\lim_{x \to 0} (b \cdot \ln a)} \]

where \( a = \frac{1^x + 2^x + \ldots + n^x}{n} \) and \( b = \frac{1}{x} \).

### Step 4: Rewrite the limit using logarithms

We can rewrite our limit as:

\[ \lim_{x \to 0} \frac{1}{x} \ln \left( \frac{1^x + 2^x + \ldots + n^x}{n} \right). \]

### Step 5: Expand using Taylor series

Using the Taylor expansion for \( k^x \) around \( x = 0 \):

\[ k^x = e^{x \ln k} \approx 1 + x \ln k \quad \text{for small } x. \]

Thus,

\[ 1^x + 2^x + \ldots + n^x \approx n + x(\ln 1 + \ln 2 + \ldots + \ln n) = n + x \ln(n!). \]

### Step 6: Substitute back into the limit

Now substituting back, we have:

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

Taking the logarithm:

\[ \ln \left( \frac{1^x + 2^x + \ldots + n^x}{n} \right) \approx \frac{x \ln(n!)}{n}. \]

### Step 7: Final limit calculation

Now substituting this back into our limit gives:

\[ \lim_{x \to 0} \frac{1}{x} \cdot \frac{x \ln(n!)}{n} = \frac{\ln(n!)}{n}. \]

Thus, we have:

\[ \lim_{x \to 0} \left( \frac{1^x + 2^x + \ldots + n^x}{n} \right)^{\frac{1}{x}} = e^{\frac{\ln(n!)}{n}} = (n!)^{\frac{1}{n}}. \]

### Final Answer

Therefore, the final result is:

\[ \lim_{x \to 0} \left( \frac{1^x + 2^x + 3^x + \ldots + n^x}{n} \right)^{\frac{1}{x}} = (n!)^{\frac{1}{n}}. \]