If `A=lim_(x to 0) (sin^(-1)(sinx))/(cos^(-1)(cosx))and B=lim_(x to 0)([|x|])/(x),` then

Options

A. A = 1
B. A does not exist
C. B = 0
D. B = 1

Explanation

To solve the problem, we need to evaluate the limits \( A \) and \( B \) as defined in the question.

### Step 1: Evaluate \( A = \lim_{x \to 0} \frac{\sin^{-1}(\sin x)}{\cos^{-1}(\cos x)} \)

1. **Understanding the Functions**:
- The function \( \sin^{-1}(\sin x) \) returns \( x \) for \( x \) in the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\). As \( x \to 0 \), this simplifies to \( \sin^{-1}(\sin x) = x \).
- The function \( \cos^{-1}(\cos x) \) returns \( x \) for \( x \) in the interval \([0, \pi]\). As \( x \to 0 \), this simplifies to \( \cos^{-1}(\cos x) = x \).

2. **Substituting in the Limit**:
\[ A = \lim_{x \to 0} \frac{x}{x} \]

3. **Simplifying the Expression**:
\[ A = \lim_{x \to 0} 1 = 1 \]

### Step 2: Evaluate \( B = \lim_{x \to 0} \frac{|x|}{x} \)

1. **Understanding the Absolute Value**:
- The expression \( \frac{|x|}{x} \) behaves differently depending on whether \( x \) approaches 0 from the positive or negative side.
- If \( x \to 0^+ \) (from the right), \( |x| = x \), so \( \frac{|x|}{x} = 1 \).
- If \( x \to 0^- \) (from the left), \( |x| = -x \), so \( \frac{|x|}{x} = -1 \).

2. **Finding the Limit**:
- Since the left-hand limit (as \( x \to 0^- \)) is -1 and the right-hand limit (as \( x \to 0^+ \)) is 1, the limit does not exist.

### Final Results:
- \( A = 1 \)
- \( B \) does not exist.

### Conclusion:
- The correct options are:
- \( A = 1 \) (Option 1)
- \( B \) does not exist (Option 2)