If `f(x)={{:(e^(|x|+|x|-1)/(|x|+|x|),":",xne0),(-1,":",x=0):}` (where `[.]` denotes the greatest integer integer function), then

Options

A. `f(x)` is continuous at x = 0
B. `lim_(xrarr0^(+))f(x)=-1`
C. `lim_(xrarr0^(-))f(x)=1`
D. `lim_(rarr0^(+))f(x)=1`

Answer

To solve the problem, we need to analyze the function \( f(x) \) given by:

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

We need to check the continuity of \( f(x) \) at \( x = 0 \). For \( f(x) \) to be continuous at \( x = 0 \), the following condition must hold:

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

### Step 1: Calculate \( f(0) \)

From the definition of the function, we have:

\[ f(0) = -1 \]

### Step 2: Calculate the Left-Hand Limit \( \lim_{x \to 0^-} f(x) \)

For \( x < 0 \), we have \( |x| = -x \). Therefore, we can rewrite the function as:

\[ f(x) = \frac{e^{-x - x - 1}}{-x - x} = \frac{e^{-2x - 1}}{-2x} \]

Now, we need to find the limit as \( x \) approaches \( 0 \) from the left:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{e^{-2x - 1}}{-2x} \]

As \( x \to 0^- \), \( e^{-2x - 1} \) approaches \( e^{-1} \) and the denominator approaches \( 0^- \), leading to:

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

### Step 3: Calculate the Right-Hand Limit \( \lim_{x \to 0^+} f(x) \)

For \( x > 0 \), we have \( |x| = x \). Therefore, we can rewrite the function as:

\[ f(x) = \frac{e^{x + x - 1}}{x + x} = \frac{e^{2x - 1}}{2x} \]

Now, we need to find the limit as \( x \) approaches \( 0 \) from the right:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{e^{2x - 1}}{2x} \]

As \( x \to 0^+ \), \( e^{2x - 1} \) approaches \( e^{-1} \) and the denominator approaches \( 0^+ \), leading to:

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

### Step 4: Conclusion

Since the left-hand limit \( \lim_{x \to 0^-} f(x) = -\infty \) and the right-hand limit \( \lim_{x \to 0^+} f(x) = +\infty \), we find that:

\[ \lim_{x \to 0} f(x) \text{ does not exist} \]

Since the limit does not exist and does not equal \( f(0) = -1 \), we conclude that the function \( f(x) \) is not continuous at \( x = 0 \).

### Final Answer

The function \( f(x) \) is not continuous at \( x = 0 \).

---