`underset(n to oo)lim ((n^(2)-n+1)/(n^(2)-n-1))^(n(n-1))`

Options

A. e
B. `e^(2)`
C. `e^(-1)`
D. 1

Answer

To solve the limit

\[ \lim_{n \to \infty} \left( \frac{n^2 - n + 1}{n^2 - n - 1} \right)^{n(n-1)}, \]

we will follow these steps:

### Step 1: Analyze the limit expression
As \( n \) approaches infinity, both the numerator and denominator approach infinity, leading to an indeterminate form of type \( \frac{\infty}{\infty} \).

### Step 2: Simplify the fraction
We can simplify the fraction inside the limit:

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

### Step 3: Take the limit of the fraction
As \( n \to \infty \):

\[ \frac{1 - \frac{1}{n} + \frac{1}{n^2}}{1 - \frac{1}{n} - \frac{1}{n^2}} \to \frac{1 - 0 + 0}{1 - 0 - 0} = 1. \]

### Step 4: Identify the form of the limit
Now we have:

\[ \lim_{n \to \infty} \left( \frac{n^2 - n + 1}{n^2 - n - 1} \right)^{n(n-1)} = 1^{\infty}, \]

which is again an indeterminate form.

### Step 5: Use logarithmic transformation
To resolve this, we can take the natural logarithm:

\[ L = \lim_{n \to \infty} n(n-1) \ln\left( \frac{n^2 - n + 1}{n^2 - n - 1} \right). \]

### Step 6: Expand the logarithm
Using the expansion of \( \ln(1 + x) \approx x \) for small \( x \):

\[ \ln\left( \frac{n^2 - n + 1}{n^2 - n - 1} \right) = \ln\left( 1 + \frac{2}{n^2 - n - 1} \right) \approx \frac{2}{n^2 - n - 1}. \]

### Step 7: Substitute back into the limit
Thus,

\[ L \approx \lim_{n \to \infty} n(n-1) \cdot \frac{2}{n^2 - n - 1}. \]

### Step 8: Simplify the expression
As \( n \to \infty \), \( n^2 - n - 1 \approx n^2 \):

\[ L \approx \lim_{n \to \infty} n(n-1) \cdot \frac{2}{n^2} = \lim_{n \to \infty} \frac{2n(n-1)}{n^2} = \lim_{n \to \infty} 2 \left( 1 - \frac{1}{n} \right) = 2. \]

### Step 9: Exponentiate to find the final limit
Thus, we have:

\[ L = 2 \implies e^L = e^2. \]

### Final Answer
Therefore, the limit is:

\[ \lim_{n \to \infty} \left( \frac{n^2 - n + 1}{n^2 - n - 1} \right)^{n(n-1)} = e^2. \]