The value of `lim_(xrarr1){(x^n-1)/(n(x-1))}^((1)/(x-1))`, is

Options

A. `e^(1//2)`
B. `e^((n)/(x-1))`
C. `e^((n-1)/(2))`
D. `e^((n+1)/(2))`

Explanation

To solve the limit \( L = \lim_{x \to 1} \left( \frac{x^n - 1}{n(x - 1)} \right)^{\frac{1}{x - 1}} \), we can follow these steps:

### Step 1: Rewrite the limit
We start by rewriting the limit in a more manageable form:
\[ L = \lim_{x \to 1} \left( \frac{x^n - 1}{n(x - 1)} \right)^{\frac{1}{x - 1}}. \]

### Step 2: Factor \( x^n - 1 \)
Using the factorization of \( x^n - 1 \), we can express it as:
\[ x^n - 1 = (x - 1)(1 + x + x^2 + \ldots + x^{n-1}). \]
Thus, we can rewrite the limit as:
\[ L = \lim_{x \to 1} \left( \frac{(x - 1)(1 + x + x^2 + \ldots + x^{n-1})}{n(x - 1)} \right)^{\frac{1}{x - 1}}. \]

### Step 3: Cancel \( x - 1 \)
Now, we can cancel \( x - 1 \) in the numerator and denominator:
\[ L = \lim_{x \to 1} \left( \frac{1 + x + x^2 + \ldots + x^{n-1}}{n} \right)^{\frac{1}{x - 1}}. \]

### Step 4: Simplify the expression
We can express the sum \( 1 + x + x^2 + \ldots + x^{n-1} \) as:
\[ 1 + x + x^2 + \ldots + x^{n-1} = \frac{x^n - 1}{x - 1}. \]
So, we have:
\[ L = \lim_{x \to 1} \left( \frac{\frac{x^n - 1}{x - 1}}{n} \right)^{\frac{1}{x - 1}}. \]

### Step 5: Rewrite using limits
Now, we can express the limit in a form suitable for applying the exponential limit:
\[ L = \lim_{x \to 1} \left( \frac{x^n - 1}{n(x - 1)} \right)^{\frac{1}{x - 1}}. \]
This can be evaluated using the fact that if \( f(x) \to 1 \) and \( g(x) \to 0 \) as \( x \to a \), then:
\[ \lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} g(x)(f(x) - 1)}. \]

### Step 6: Calculate \( f(x) - 1 \)
We find \( f(x) - 1 \):
\[ f(x) = \frac{x^n - 1}{n(x - 1)} \quad \text{and} \quad f(1) = \frac{0}{0} \text{ (indeterminate form)}. \]
Using L'Hôpital's rule or Taylor expansion around \( x = 1 \), we find:
\[ f(x) - 1 \approx \frac{1 + 2 + 3 + \ldots + (n-1)}{n} \cdot (x - 1). \]

### Step 7: Evaluate the limit
Now, we can evaluate:
\[ L = e^{\lim_{x \to 1} \frac{1}{x - 1} \cdot \left( \frac{n(n-1)}{2n} (x - 1) \right)} = e^{\frac{(n-1)}{2}}. \]

### Final Result
Thus, the final result is:
\[ L = e^{\frac{n - 1}{2}}. \]