The value of `lim_(x to oo ) (x-x^(2)log_(e)(1+(1)/(x)))` is _______.

Options

A. 1
B. `(1)/(2)`
C. `(1)/(4)`
D. 2

Explanation

To solve the limit \( \lim_{x \to \infty} \left( x - x^2 \log_e\left(1 + \frac{1}{x}\right) \right) \), we can follow these steps:

### Step 1: Analyze the limit
We start by substituting \( x \) directly into the limit:
\[ \lim_{x \to \infty} \left( x - x^2 \log_e\left(1 + \frac{1}{x}\right) \right) \]
As \( x \) approaches infinity, \( \frac{1}{x} \) approaches 0, so we have:
\[ \log_e\left(1 + \frac{1}{x}\right) \to \log_e(1) = 0 \]
Thus, we have an indeterminate form of \( \infty - \infty \).

### Step 2: Rewrite the limit
To resolve the indeterminate form, we can rewrite the limit:
\[ x - x^2 \log_e\left(1 + \frac{1}{x}\right) = x \left(1 - x \log_e\left(1 + \frac{1}{x}\right)\right) \]
Now we need to analyze \( x \log_e\left(1 + \frac{1}{x}\right) \) as \( x \to \infty \).

### Step 3: Use Taylor expansion
Using the Taylor series expansion for \( \log_e(1 + u) \) around \( u = 0 \):
\[ \log_e(1 + u) \approx u - \frac{u^2}{2} + O(u^3) \]
Substituting \( u = \frac{1}{x} \):
\[ \log_e\left(1 + \frac{1}{x}\right) \approx \frac{1}{x} - \frac{1}{2x^2} + O\left(\frac{1}{x^3}\right) \]
Thus,
\[ x \log_e\left(1 + \frac{1}{x}\right) \approx 1 - \frac{1}{2x} + O\left(\frac{1}{x^2}\right) \]

### Step 4: Substitute back into the limit
Now substituting back into our limit:
\[ 1 - x \log_e\left(1 + \frac{1}{x}\right) \approx 1 - \left(1 - \frac{1}{2x} + O\left(\frac{1}{x^2}\right)\right) = \frac{1}{2x} - O\left(\frac{1}{x^2}\right) \]
So we have:
\[ x \left(1 - x \log_e\left(1 + \frac{1}{x}\right)\right) \approx x \cdot \left(\frac{1}{2x} - O\left(\frac{1}{x^2}\right)\right) = \frac{1}{2} - O\left(\frac{1}{x}\right) \]

### Step 5: Take the limit
Taking the limit as \( x \to \infty \):
\[ \lim_{x \to \infty} \left(\frac{1}{2} - O\left(\frac{1}{x}\right)\right) = \frac{1}{2} \]

### Final Answer
Thus, the value of the limit is:
\[ \frac{1}{2} \]