If `f(x)={{:(,(e^(x[x])-1)/(x+[x]),x ne 0),(,1,x=0):}` then

Options

A. `underset(x to 0^(+))lim f(x)=-1`
B. `underset(x to 0^(-))lim f(x)=(1)/(e)-1`
C. f(x) is continuous at x=0
D. f(x) is discontinuous at x=0

Explanation

To solve the problem, we need to analyze the function given:

\[ f(x) = \begin{cases} \frac{e^{x[x]} - 1}{x + [x]} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \]

where \([x]\) denotes the greatest integer function (or floor function) of \(x\).

### Step 1: Find the Left-Hand Limit (LHL) as \(x\) approaches 0 from the left
For \(x < 0\), \([x] = -1\) (since the greatest integer less than any negative number is one less than that number). Thus, we can rewrite \(f(x)\):

\[ f(x) = \frac{e^{x(-1)} - 1}{x - 1} = \frac{e^{-x} - 1}{x - 1} \]

Now, we need to find:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{e^{-x} - 1}{x - 1} \]

### Step 2: Calculate the limit
As \(x\) approaches 0 from the left:

\[ \lim_{x \to 0^-} f(x) = \frac{e^{0} - 1}{0 - 1} = \frac{1 - 1}{-1} = \frac{0}{-1} = 0 \]

### Step 3: Find the Right-Hand Limit (RHL) as \(x\) approaches 0 from the right
For \(x > 0\), \([x] = 0\). Thus, we can rewrite \(f(x)\):

\[ f(x) = \frac{e^{x(0)} - 1}{x + 0} = \frac{1 - 1}{x} = 0 \]

Now, we need to find:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{e^{0} - 1}{x} = \frac{1 - 1}{0} = 0 \]

### Step 4: Check the value of \(f(0)\)
From the definition of the function, we know:

\[ f(0) = 1 \]

### Step 5: Determine continuity at \(x = 0\)
For \(f(x)\) to be continuous at \(x = 0\), the following must hold:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]

From our calculations:

- \(\lim_{x \to 0^-} f(x) = 0\)
- \(\lim_{x \to 0^+} f(x) = 0\)
- \(f(0) = 1\)

Since \(0 \neq 1\), the function \(f(x)\) is discontinuous at \(x = 0\).

### Conclusion
Thus, the correct option is that \(f(x)\) is discontinuous at \(x = 0\).

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