`lim_(xrarr0)((1^(x)+2^(x)+3^(x)+…..+n^(x))/(n))^((a)/(x))` equals

Options

A. `(n!)^((a)/(n))`
B. `(n!)^(an)`
C. None of these
D. `(n!)^((n)/(a))`

Answer

To solve the limit

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

we will follow these steps:

### Step 1: Rewrite the expression
We start by rewriting the expression inside the limit:

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

### Step 2: Evaluate the limit of the numerator
As \( x \to 0 \), each term \( k^x \) (for \( k = 1, 2, \ldots, n \)) approaches 1. Therefore, we can express the limit as:

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

### Step 3: Recognize the indeterminate form
Now we have an expression of the form \( 1^{\infty} \). To resolve this, we can use the logarithmic limit approach:

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

### Step 4: Simplify the expression inside the limit
We need to analyze the expression:

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

### Step 5: Use Taylor expansion
Using the Taylor expansion for \( k^x \) around \( x = 0 \):

\[ k^x \approx 1 + x \ln k, \]

we can write:

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

Thus, we have:

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

### Step 6: Substitute back into the limit
Now substituting back into our limit, we get:

\[ L = e^{\lim_{x \to 0} \frac{a}{x} \cdot \frac{x \ln(n!)}{n}} = e^{\frac{a \ln(n!)}{n}}. \]

### Step 7: Final expression
Thus, our final result for the limit is:

\[ L = e^{\frac{a \ln(n!)}{n}} = (n!)^{\frac{a}{n}}. \]

### Final Answer
The limit evaluates to:

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

---