`lim_(x->0) [1+[x]]^(2/x),` where [:] is greatest integer function, is equal to

Options

A. 0
B. 1
C. `e ^(2)`
D. Does not exist

Answer

To solve the limit \( \lim_{x \to 0} \left[ 1 + [x] \right]^{\frac{2}{x}} \), where \([x]\) is the greatest integer function (also known as the floor function), we will analyze the behavior of the function as \(x\) approaches 0 from both the left and the right.

### Step-by-step Solution:

1. **Understanding the Greatest Integer Function**:
- The greatest integer function \([x]\) gives the largest integer less than or equal to \(x\).
- As \(x\) approaches 0 from the left (i.e., \(x \to 0^{-}\)), \([x] = -1\).
- As \(x\) approaches 0 from the right (i.e., \(x \to 0^{+}\)), \([x] = 0\).

2. **Finding the Left-Hand Limit (LHL)**:
- For \(x \to 0^{-}\):
\[ \lim_{x \to 0^{-}} [1 + [x]]^{\frac{2}{x}} = \lim_{x \to 0^{-}} [1 - 1]^{\frac{2}{x}} = \lim_{x \to 0^{-}} 0^{\frac{2}{x}}. \]
- Since \(0^{\frac{2}{x}}\) approaches 0 for any negative \(x\), we have:
\[ \text{LHL} = 0. \]

3. **Finding the Right-Hand Limit (RHL)**:
- For \(x \to 0^{+}\):
\[ \lim_{x \to 0^{+}} [1 + [x]]^{\frac{2}{x}} = \lim_{x \to 0^{+}} [1 + 0]^{\frac{2}{x}} = \lim_{x \to 0^{+}} 1^{\frac{2}{x}}. \]
- Since \(1^{\frac{2}{x}} = 1\) for any \(x\), we have:
\[ \text{RHL} = 1. \]

4. **Conclusion**:
- Since the left-hand limit (0) is not equal to the right-hand limit (1), the overall limit does not exist:
\[ \lim_{x \to 0} \left[ 1 + [x] \right]^{\frac{2}{x}} \text{ does not exist.} \]

### Final Answer:
The limit does not exist.