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 \( f(x) \) defined as:

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

We are particularly interested in the behavior of \( f(x) \) as \( x \) approaches 0 from both the left and the right.

### Step 1: Evaluate \( \lim_{x \to 0^-} f(x) \)

For \( x \to 0^- \) (approaching from the left), we have \( |x| = -x \). Thus, we can rewrite \( f(x) \):

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

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 \( -2x \) approaches 0. Therefore, we can apply L'Hôpital's Rule since we have an indeterminate form of type \( \frac{0}{0} \).

Taking the derivative of the numerator and denominator:

\[ \text{Numerator's derivative: } \frac{d}{dx}(e^{-2x - 1}) = -2e^{-2x - 1} \]
\[ \text{Denominator's derivative: } \frac{d}{dx}(-2x) = -2 \]

Now applying L'Hôpital's Rule:

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

### Step 2: Evaluate \( \lim_{x \to 0^+} f(x) \)

For \( x \to 0^+ \) (approaching from the right), we have \( |x| = x \). Thus, we can rewrite \( f(x) \):

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

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 \( 2x \) approaches 0. Again, we have an indeterminate form \( \frac{0}{0} \), so we apply L'Hôpital's Rule:

Taking the derivative of the numerator and denominator:

\[ \text{Numerator's derivative: } \frac{d}{dx}(e^{2x - 1}) = 2e^{2x - 1} \]
\[ \text{Denominator's derivative: } \frac{d}{dx}(2x) = 2 \]

Now applying L'Hôpital's Rule:

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

### Step 3: Compare the limits and check continuity

We have:

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

Both limits are equal, but we also need to check the value of \( f(0) \):

\[ f(0) = -1 \]

Since the limits from both sides do not equal \( f(0) \), the function is not continuous at \( x = 0 \).

### Conclusion

The only correct statement is that \( \lim_{x \to 0^-} f(x) = e^{-1} \).