which of the following function(s) not defined at `x=0` has/have removable discontinuity at `x = 0`.

Options

A. `f(x) = (1)/(1+2^(cot x))`
B. `f(x) = "cos"(((|sin x|))/(x))`
C. `f(x) = "x sin"(pi)/(x)`
D. `f(x) = (1)/("In"|x|)`

Answer

To determine which of the given functions have removable discontinuity at \( x = 0 \), we need to analyze each function based on the definition of removable discontinuity. A function \( f(x) \) has a removable discontinuity at \( x = a \) if:

1. The limits \( \lim_{x \to a^-} f(x) \) and \( \lim_{x \to a^+} f(x) \) exist and are equal.
2. The limit is not equal to \( f(a) \) (or \( f(a) \) is not defined).

Let's analyze each function step by step.

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

1. **Left-Hand Limit (LHL) as \( x \to 0^- \)**:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1}{1 + 2^{\cot x}} = \frac{1}{1 + 2^{\infty}} = \frac{1}{1 + \infty} = 0 \]

2. **Right-Hand Limit (RHL) as \( x \to 0^+ \)**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1}{1 + 2^{\cot x}} = \frac{1}{1 + 2^{\infty}} = \frac{1}{1 + \infty} = 0 \]

3. **Conclusion**:
- LHL = 0
- RHL = 0
- Since \( f(0) \) is not defined, we have a removable discontinuity.

### Function 2: \( f(x) = \frac{\cos(\lvert \sin x \rvert)}{x} \)

1. **Left-Hand Limit (LHL) as \( x \to 0^- \)**:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\cos(\lvert \sin x \rvert)}{x} = \frac{\cos(0)}{0} = \text{undefined} \]

2. **Right-Hand Limit (RHL) as \( x \to 0^+ \)**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\cos(\lvert \sin x \rvert)}{x} = \frac{\cos(0)}{0} = \text{undefined} \]

3. **Conclusion**:
- Both limits are undefined, hence no removable discontinuity.

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

1. **Left-Hand Limit (LHL) as \( x \to 0^- \)**:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x \sin\left(\frac{\pi}{x}\right) = 0 \cdot \text{(bounded)} = 0 \]

2. **Right-Hand Limit (RHL) as \( x \to 0^+ \)**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x \sin\left(\frac{\pi}{x}\right) = 0 \cdot \text{(bounded)} = 0 \]

3. **Conclusion**:
- LHL = 0
- RHL = 0
- Since \( f(0) = 0 \), this is a continuous function, not a removable discontinuity.

### Function 4: \( f(x) = \frac{1}{\ln(\lvert x \rvert)} \)

1. **Left-Hand Limit (LHL) as \( x \to 0^- \)**:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1}{\ln(\lvert x \rvert)} = \frac{1}{-\infty} = 0 \]

2. **Right-Hand Limit (RHL) as \( x \to 0^+ \)**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1}{\ln(\lvert x \rvert)} = \frac{1}{-\infty} = 0 \]

3. **Conclusion**:
- LHL = 0
- RHL = 0
- Since \( f(0) \) is not defined, we have a removable discontinuity.

### Final Conclusion
The only function that has a removable discontinuity at \( x = 0 \) is:

- **Function 1: \( f(x) = \frac{1}{1 + 2^{\cot x}} \)**
- **Function 4: \( f(x) = \frac{1}{\ln(\lvert x \rvert)} \)**