`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 logarithm
We start by rewriting the logarithmic term:
\[ \log_e \left( \frac{e^x + e^{-x}}{2} \right) = \log_e (e^x + e^{-x}) - \log_e(2) \]
Thus, we can express the limit as:
\[ L = \lim_{x \to \infty} \left[ x - \log_e (e^x + e^{-x}) + \log_e(2) \right] \]

### Step 2: Simplify \( e^x + e^{-x} \)
As \( x \to \infty \), \( e^{-x} \) approaches 0, so:
\[ e^x + e^{-x} \approx e^x \]
Thus, we can approximate:
\[ \log_e (e^x + e^{-x}) \approx \log_e (e^x) = x \]

### Step 3: Substitute back into the limit
Substituting this approximation back into our limit gives:
\[ L = \lim_{x \to \infty} \left[ x - x + \log_e(2) \right] \]
This simplifies to:
\[ L = \lim_{x \to \infty} \log_e(2) = \log_e(2) \]

### Final Result
Thus, the limit evaluates to:
\[ \boxed{\log_e(2)} \]

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