If `f(x)={{:(,|x|+1, x lt 0),(, 0,x=0),(,|x|-1, x gt 0):}" then "underset(x to a)lim` f(x) exists for all

Options

A. `a ne 1`
B. `a ne 0`
C. `a ne -1`
D. `a ne 2`

Answer

To determine the values of \( a \) for which the limit \( \lim_{x \to a} f(x) \) exists, we need to analyze the function \( f(x) \) defined as follows:

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

### Step 1: Identify the function behavior around critical points
We need to check the limit as \( x \) approaches 0, since this is where the function definition changes.

### Step 2: Calculate the left-hand limit as \( x \to 0^- \)
For \( x < 0 \), the function is defined as \( f(x) = |x| + 1 = -x + 1 \). Thus, we compute:

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

### Step 3: Calculate the right-hand limit as \( x \to 0^+ \)
For \( x > 0 \), the function is defined as \( f(x) = |x| - 1 = x - 1 \). Thus, we compute:

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

### Step 4: Compare the left-hand and right-hand limits
We found:

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

Since the left-hand limit does not equal the right-hand limit, we conclude that:

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

### Step 5: Determine the values of \( a \) for which the limit exists
The limit \( \lim_{x \to a} f(x) \) exists for all \( a \) except where the function is discontinuous. We found that the limit does not exist at \( a = 0 \).

### Conclusion
Thus, the limit exists for all \( a \) except \( a = 0 \). Therefore, the final answer is:

\[ \text{The limit } \lim_{x \to a} f(x) \text{ exists for all } a \neq 0. \]