Let `f(x)={{:("x-|x|",","x ne0),(1,","x=0):}`then

Options

A. `lim_(xto0+) f(x)=1`
B. `lim_(x to 0-) f(x)=0`
C. `lim_(x to 0+) f(x) ne lim_(x to 0-) f(x)`
D. `lim_(x to 0+) f(x)` does not exist

Answer

To solve the problem, we need to analyze the function defined as:

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

We will find the left-hand limit and the right-hand limit as \( x \) approaches 0.

### Step 1: Determine the left-hand limit as \( x \) approaches 0

For \( x < 0 \):
- The expression for \( |x| \) is \( -x \).
- Therefore, \( f(x) = x - (-x) = x + x = 2x \).

Now we calculate the left-hand limit:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 2x = 2 \cdot 0 = 0. \]

### Step 2: Determine the right-hand limit as \( x \) approaches 0

For \( x > 0 \):
- The expression for \( |x| \) is \( x \).
- Therefore, \( f(x) = x - x = 0 \).

Now we calculate the right-hand limit:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 0 = 0. \]

### Step 3: Compare the left-hand limit and right-hand limit

From the calculations:
- Left-hand limit: \( \lim_{x \to 0^-} f(x) = 0 \)
- Right-hand limit: \( \lim_{x \to 0^+} f(x) = 0 \)

Since both limits are equal, we can conclude:
\[ \lim_{x \to 0} f(x) = 0. \]

### Step 4: Evaluate the function at \( x = 0 \)

According to the definition of the function:
\[ f(0) = 1. \]

### Conclusion

The limit as \( x \) approaches 0 exists and is equal to 0, but the value of the function at \( x = 0 \) is 1. Therefore, the function is not continuous at \( x = 0 \).

### Summary of Results
- \( \lim_{x \to 0^-} f(x) = 0 \)
- \( \lim_{x \to 0^+} f(x) = 0 \)
- \( f(0) = 1 \)