Let `f(x)={(x(e^([x]+x)-4)/([x]+|x|), , x!=0),(3, , x=0):}`
Where [ ] denotes the greatest integer function. Then,

Options

A. f(x) is discontinuous at x = 0
B. f (x) is continuous at x = 0
C. `lim_(x to 0)f(x)` exists
D. None of these

Answer

To determine the continuity of the function \( f(x) \) at \( x = 0 \), we need to analyze the left-hand limit, right-hand limit, and the value of the function at that point.

Given the function:
\[ f(x) = \begin{cases} \frac{x(e^{[x] + x} - 4)}{[x] + |x|} & \text{if } x \neq 0 \\ 3 & \text{if } x = 0 \end{cases} \]
where \([x]\) denotes the greatest integer function.

### Step 1: Evaluate \( f(0) \)
The value of the function at \( x = 0 \) is given as:
\[ f(0) = 3 \]

### Step 2: Calculate the right-hand limit as \( x \to 0^+ \)
For \( x \to 0^+ \):
- \([x] = 0\)
- \(|x| = x\)

Thus, the expression simplifies to:
\[ f(x) = \frac{x(e^{0 + x} - 4)}{0 + x} = \frac{x(e^{x} - 4)}{x} = e^{x} - 4 \]
Now, we find the limit:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (e^{x} - 4) = e^{0} - 4 = 1 - 4 = -3 \]

### Step 3: Calculate the left-hand limit as \( x \to 0^- \)
For \( x \to 0^- \):
- \([x] = -1\)
- \(|x| = -x\)

Thus, the expression simplifies to:
\[ f(x) = \frac{x(e^{-1 + x} - 4)}{-1 - x} = \frac{x(e^{-1 + x} - 4)}{-1 - x} \]
As \( x \to 0^- \), we can evaluate the limit:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x(e^{-1} - 4)}{-1} = \frac{0 \cdot (e^{-1} - 4)}{-1} = 0 \]

### Step 4: Compare limits and function value
We have:
- \( \lim_{x \to 0^+} f(x) = -3 \)
- \( \lim_{x \to 0^-} f(x) = 0 \)
- \( f(0) = 3 \)

Since the left-hand limit and right-hand limit are not equal, and neither is equal to \( f(0) \), we conclude that \( f(x) \) is discontinuous at \( x = 0 \).

### Conclusion
The function \( f(x) \) is discontinuous at \( x = 0 \).