Let f(x)=sgn (sgn (sgn x)). Then `lim_(x to0)` f(x) is

Options

A. 1
B. 2
C. 0
D. none of these

Answer

To solve the problem, we need to evaluate the limit of the function \( f(x) = \text{sgn}(\text{sgn}(\text{sgn}(x))) \) as \( x \) approaches 0. The signum function, \( \text{sgn}(x) \), is defined as follows:

- \( \text{sgn}(x) = 1 \) if \( x > 0 \)
- \( \text{sgn}(x) = 0 \) if \( x = 0 \)
- \( \text{sgn}(x) = -1 \) if \( x < 0 \)

### Step-by-Step Solution:

1. **Evaluate the Left-Hand Limit (LHL)**:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \text{sgn}(\text{sgn}(\text{sgn}(x))) \]
For \( x \to 0^- \) (approaching from the left), \( x \) is negative, so:
\[ \text{sgn}(x) = -1 \]
Now we substitute this value into the next signum function:
\[ \text{sgn}(-1) = -1 \]
Finally, we substitute again:
\[ \text{sgn}(-1) = -1 \]
Thus, we find:
\[ \lim_{x \to 0^-} f(x) = -1 \]

2. **Evaluate the Right-Hand Limit (RHL)**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \text{sgn}(\text{sgn}(\text{sgn}(x))) \]
For \( x \to 0^+ \) (approaching from the right), \( x \) is positive, so:
\[ \text{sgn}(x) = 1 \]
Now we substitute this value into the next signum function:
\[ \text{sgn}(1) = 1 \]
Finally, we substitute again:
\[ \text{sgn}(1) = 1 \]
Thus, we find:
\[ \lim_{x \to 0^+} f(x) = 1 \]

3. **Conclusion**:
Since the left-hand limit and right-hand limit are not equal:
\[ \lim_{x \to 0^-} f(x) = -1 \quad \text{and} \quad \lim_{x \to 0^+} f(x) = 1 \]
Therefore, the limit \( \lim_{x \to 0} f(x) \) does not exist.

### Final Answer:
The limit \( \lim_{x \to 0} f(x) \) does not exist.