For which of the following functions `f(0)` exists such that `f(x)` is continuous at `f(x)=1/((log)_e|x|)` b. `f(x)=1/((log)_"e"|x|)` c. f(x)=x sinpi/x d. `f(x)=1/(1+2^(cot x))`

Options

A. `f(x)=(1)/(log_(e)|x|)`
B. `f(x)=cos((|sinx|)/(x))`
C. `f(x)=x sin (pi)/(x)`
D. `f(x)=(1)=(1)/(1+2^(cotx))`

Answer

To determine for which of the given functions \( f(0) \) exists such that \( f(x) \) is continuous at \( x = 0 \), we will analyze each function one by one.

### Given Functions:
1. \( f(x) = \frac{1}{\log_e |x|} \)
2. \( f(x) = \frac{\cos(|\sin x|)}{x} \)
3. \( f(x) = x \sin\left(\frac{\pi}{x}\right) \)
4. \( f(x) = \frac{1}{1 + 2^{\cot x}} \)

### Step-by-Step Analysis:

#### 1. Function: \( f(x) = \frac{1}{\log_e |x|} \)

- **Check \( f(0) \)**:
\[ f(0) = \frac{1}{\log_e |0|} \quad \text{(undefined since } \log_e 0 \text{ does not exist)} \]
- **Conclusion**: \( f(0) \) does not exist. **Not continuous at \( x = 0 \)**.

#### 2. Function: \( f(x) = \frac{\cos(|\sin x|)}{x} \)

- **Check \( f(0) \)**:
\[ f(0) = \frac{\cos(|\sin 0|)}{0} = \frac{\cos(0)}{0} = \frac{1}{0} \quad \text{(undefined)} \]
- **Conclusion**: \( f(0) \) does not exist. **Not continuous at \( x = 0 \)**.

#### 3. Function: \( f(x) = x \sin\left(\frac{\pi}{x}\right) \)

- **Check \( f(0) \)**:
\[ f(0) = 0 \cdot \sin\left(\frac{\pi}{0}\right) \quad \text{(undefined)} \]
- However, we need to check the limit as \( x \to 0 \):
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x \sin\left(\frac{\pi}{x}\right) \]
- Since \( \sin\left(\frac{\pi}{x}\right) \) oscillates between -1 and 1:
\[ -|x| \leq x \sin\left(\frac{\pi}{x}\right) \leq |x| \]
- By the Squeeze Theorem:
\[ \lim_{x \to 0} f(x) = 0 \]
- **Set \( f(0) = 0 \)** to make it continuous at \( x = 0 \).
- **Conclusion**: \( f(0) \) can be defined as 0. **Continuous at \( x = 0 \)**.

#### 4. Function: \( f(x) = \frac{1}{1 + 2^{\cot x}} \)

- **Check \( f(0) \)**:
- \( \cot x \) is undefined at \( x = 0 \), leading to:
\[ f(0) = \frac{1}{1 + 2^{\cot 0}} \quad \text{(undefined)} \]
- **Conclusion**: \( f(0) \) does not exist. **Not continuous at \( x = 0 \)**.

### Final Conclusion:
The only function for which \( f(0) \) exists and \( f(x) \) is continuous at \( x = 0 \) is:
- **Option 3: \( f(x) = x \sin\left(\frac{\pi}{x}\right) \)**