`lim_(xrarroo) [x-log_(e)((e^(x)+e^(-x))/(2))]=`

Options

A. `log_(e)4`
B. 0
C. `log_(e)2`
D. `oo`

Explanation

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

### Step 1: Rewrite the logarithmic expression
We start with the expression inside the limit:
\[ \lim_{x \to \infty} \left[ x - \log_e\left(\frac{e^x + e^{-x}}{2}\right) \right] \]
We can rewrite the logarithm:
\[ \log_e\left(\frac{e^x + e^{-x}}{2}\right) = \log_e(e^x + e^{-x}) - \log_e(2) \]
Thus, the limit becomes:
\[ \lim_{x \to \infty} \left[ x - \log_e(e^x + e^{-x}) + \log_e(2) \right] \]

### Step 2: Simplify \( \log_e(e^x + e^{-x}) \)
As \( x \to \infty \), \( e^{-x} \) approaches 0, so:
\[ e^x + e^{-x} \approx e^x \]
Thus:
\[ \log_e(e^x + e^{-x}) \approx \log_e(e^x) = x \]
Now substituting this back into our limit:
\[ \lim_{x \to \infty} \left[ x - x + \log_e(2) \right] \]

### Step 3: Evaluate the limit
The expression simplifies to:
\[ \lim_{x \to \infty} \left[ 0 + \log_e(2) \right] = \log_e(2) \]

### Final Result
Therefore, the limit is:
\[ \lim_{x \to \infty} \left[ x - \log_e\left(\frac{e^x + e^{-x}}{2}\right) \right] = \log_e(2) \]