If both `Lim_(xrarrc^(-))f(x)` and `Lim_(xrarrc^(+))f(x)` exist finitely and are equal, then the function `f` is said to have removable discontinuity at `x=c`. If both the limits i.e. `Lim_(xrarrc^(-))f(x)` and `Lim_(xrarrc^(+))f(x)` exist finitely and are not equal, then the function `f` is said to have non-removable discontinuity at `x=c`.
Which of the following function not defined at `x=0` has removable discontinuity at the origin?

Options

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

Explanation

To determine which function has removable discontinuity at the origin (x = 0), we will analyze each function provided in the question step by step.

### Step 1: Analyze the first function
**Function:** \( f(x) = \frac{1}{1 + 2^{\cot x}} \)

- **At \( x = 0 \):** The cotangent function \( \cot x \) approaches infinity as \( x \) approaches 0, making \( f(x) \) undefined at this point.

- **Left-hand limit:**
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1}{1 + 2^{\cot x}} \to \frac{1}{1 + 0} = 1 \]

- **Right-hand limit:**
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1}{1 + 2^{\cot x}} \to \frac{1}{1 + 0} = 1 \]

Since both limits exist and are equal, this function has removable discontinuity at \( x = 0 \).

### Step 2: Analyze the second function
**Function:** \( f(x) = x \cdot \sin\left(\frac{\pi}{x}\right) \)

- **At \( x = 0 \):** The function is not defined at \( x = 0 \).

- **Left-hand limit:**
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x \cdot \sin\left(\frac{\pi}{x}\right) = 0 \]

- **Right-hand limit:**
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x \cdot \sin\left(\frac{\pi}{x}\right) = 0 \]

Both limits exist and are equal, indicating that this function also has removable discontinuity at \( x = 0 \).

### Step 3: Analyze the third function
**Function:** \( f(x) = \frac{1}{\ln|x|} \)

- **At \( x = 0 \):** The function is not defined at \( x = 0 \).

- **Left-hand limit:**
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1}{\ln(-x)} \to 0 \]

- **Right-hand limit:**
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1}{\ln(x)} \to 0 \]

Both limits exist and are equal, indicating that this function also has removable discontinuity at \( x = 0 \).

### Step 4: Analyze the fourth function
**Function:** \( f(x) = \frac{\sin(\sin|x|)}{x} \)

- **At \( x = 0 \):** The function is not defined at \( x = 0 \).

- **Left-hand limit:**
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin(\sin(-x))}{-x} = -\lim_{x \to 0} \frac{\sin(\sin(-x))}{x} \to -\sin(0) = 0 \]

- **Right-hand limit:**
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sin(\sin(x))}{x} \to \sin(0) = 0 \]

Both limits exist but are not equal, indicating that this function does not have removable discontinuity at \( x = 0 \).

### Conclusion
The functions that have removable discontinuity at the origin are:
1. \( f(x) = \frac{1}{1 + 2^{\cot x}} \)
2. \( f(x) = x \cdot \sin\left(\frac{\pi}{x}\right) \)
3. \( f(x) = \frac{1}{\ln|x|} \)

The function that does not have removable discontinuity is:
4. \( f(x) = \frac{\sin(\sin|x|)}{x} \)