If `f(x) = (e^([x] + |x|) -3)/([x] + |x|+ 1)` , then: (where [.] represents greatest integer function)

Options

A. `lim_(x to 0^(+)) f(x) = -2`
B. `lim_(x to 0^(+)) f(x) = 0`
C. `lim_(x to 0^(+)) f(x) = 2`
D. `lim_(x to 0^(+)) f(x)` exist

Explanation

To solve the problem, we need to find the right-hand limit (RHL) and left-hand limit (LHL) of the function \( f(x) = \frac{e^{[x] + |x|} - 3}{[x] + |x| + 1} \) as \( x \) approaches 0. Here, \([x]\) represents the greatest integer function (floor function), and \(|x|\) represents the absolute value of \( x \).

### Step 1: Calculate the Right-Hand Limit (RHL) as \( x \to 0^+ \)

1. **Substitute \( x \to 0^+ \)**:
- When \( x \) approaches 0 from the right, \([x] = 0\) and \(|x| = x\).
- Thus, the function becomes:
\[ f(x) = \frac{e^{0 + x} - 3}{0 + x + 1} = \frac{e^x - 3}{x + 1} \]

2. **Evaluate the limit**:
- Now, we need to find:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{e^x - 3}{x + 1} \]
- Substituting \( x = 0 \):
\[ = \frac{e^0 - 3}{0 + 1} = \frac{1 - 3}{1} = -2 \]

### Step 2: Calculate the Left-Hand Limit (LHL) as \( x \to 0^- \)

1. **Substitute \( x \to 0^- \)**:
- When \( x \) approaches 0 from the left, \([x] = -1\) and \(|x| = -x\).
- Thus, the function becomes:
\[ f(x) = \frac{e^{-1 - (-x)} - 3}{-1 + (-x) + 1} = \frac{e^{-1 + x} - 3}{-x} \]

2. **Evaluate the limit**:
- Now, we need to find:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{e^{-1 + x} - 3}{-x} \]
- Substituting \( x = 0 \):
\[ = \frac{e^{-1} - 3}{0} \text{ (This approaches infinity)} \]

### Step 3: Compare RHL and LHL

- From the calculations:
- RHL as \( x \to 0^+ \) is \(-2\).
- LHL as \( x \to 0^- \) approaches infinity.

Since the RHL and LHL are not equal, the limit does not exist at \( x = 0 \).

### Conclusion

The function \( f(x) \) does not exist at \( x = 0 \) since the left-hand limit and right-hand limit are not equal.