`lim_(xrarroo) x^(2)sin(log_(e)sqrt(cos(pi)/(x)))`

Options

A. 0
B. `-(pi^(2))/(2)`
C. `-(pi^(2))/(4)`
D. `-(pi^(2))/(8)`

Explanation

To solve the limit \( \lim_{x \to \infty} x^2 \sin\left(\log_e\sqrt{\cos\left(\frac{\pi}{x}\right)}\right) \), we will follow these steps:

### Step 1: Analyze the limit
First, we substitute \( x \to \infty \) into the expression:
\[ \cos\left(\frac{\pi}{x}\right) \to \cos(0) = 1 \]
Thus,
\[ \sqrt{\cos\left(\frac{\pi}{x}\right)} \to \sqrt{1} = 1 \]
and
\[ \log_e(1) = 0. \]
This gives us the form \( x^2 \sin(0) = 0 \cdot \infty \), which is indeterminate.

### Step 2: Rewrite the limit
To resolve the indeterminate form, we can rewrite the limit:
\[ L = \lim_{x \to \infty} x^2 \sin\left(\log_e\sqrt{\cos\left(\frac{\pi}{x}\right)}\right). \]
We know that \( \sin(h) \approx h \) when \( h \) is close to 0. Therefore, we can express:
\[ \sin\left(\log_e\sqrt{\cos\left(\frac{\pi}{x}\right)}\right) \approx \log_e\sqrt{\cos\left(\frac{\pi}{x}\right)}. \]

### Step 3: Simplify the logarithmic term
Now we simplify \( \log_e\sqrt{\cos\left(\frac{\pi}{x}\right)} \):
\[ \log_e\sqrt{\cos\left(\frac{\pi}{x}\right)} = \frac{1}{2} \log_e\left(\cos\left(\frac{\pi}{x}\right)\right). \]
Thus,
\[ L = \lim_{x \to \infty} x^2 \cdot \frac{1}{2} \log_e\left(\cos\left(\frac{\pi}{x}\right)\right). \]

### Step 4: Use Taylor expansion for cosine
Using the Taylor series expansion for \( \cos\left(\frac{\pi}{x}\right) \) around 0:
\[ \cos\left(\frac{\pi}{x}\right) \approx 1 - \frac{1}{2}\left(\frac{\pi}{x}\right)^2. \]
Then,
\[ \log_e\left(\cos\left(\frac{\pi}{x}\right)\right) \approx \log_e\left(1 - \frac{1}{2}\left(\frac{\pi}{x}\right)^2\right) \approx -\frac{1}{2}\left(\frac{\pi}{x}\right)^2. \]

### Step 5: Substitute back into the limit
Substituting this back into the limit gives:
\[ L = \lim_{x \to \infty} x^2 \cdot \frac{1}{2} \left(-\frac{1}{2}\left(\frac{\pi}{x}\right)^2\right) = \lim_{x \to \infty} -\frac{\pi^2}{4}. \]

### Step 6: Evaluate the limit
Since the limit does not depend on \( x \) anymore, we find:
\[ L = -\frac{\pi^2}{4}. \]

### Final Answer
Thus, the final answer is:
\[ \boxed{-\frac{\pi^2}{4}}. \]