Consider the following in respect of the function
`f(x)={{:(2+x","xge0),(2-x","xlt0):}`
1. `lim_(x to 1) f(x)` does not exist.
2. f(x) is differentiable at x=0
3. f(x) is continuous at x=0
Which of the above statements is /aer correct?

Options

A. 1 only
B. 3 olny
C. 2 and 3 only
D. 1 and 3 only

Explanation

To solve the problem, we will analyze the function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} 2 + x & \text{if } x \geq 0 \\ 2 - x & \text{if } x < 0 \end{cases} \]

We need to evaluate the three statements regarding the function:

1. \( \lim_{x \to 1} f(x) \) does not exist.
2. \( f(x) \) is differentiable at \( x = 0 \).
3. \( f(x) \) is continuous at \( x = 0 \).

### Step 1: Evaluate the limit as \( x \to 1 \)

Since \( 1 \geq 0 \), we will use the first case of the function:

\[ f(1) = 2 + 1 = 3 \]

Now, we find the limit:

\[ \lim_{x \to 1} f(x) = f(1) = 3 \]

**Conclusion for Statement 1**: The limit exists, hence the statement is **incorrect**.

### Step 2: Check continuity at \( x = 0 \)

To check continuity at \( x = 0 \), we need to evaluate:

- \( f(0) \)
- \( \lim_{x \to 0^-} f(x) \) (left-hand limit)
- \( \lim_{x \to 0^+} f(x) \) (right-hand limit)

Calculating \( f(0) \):

\[ f(0) = 2 + 0 = 2 \]

Calculating the left-hand limit as \( x \to 0 \) (using \( f(x) = 2 - x \) for \( x < 0 \)):

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

Calculating the right-hand limit as \( x \to 0 \) (using \( f(x) = 2 + x \) for \( x \geq 0 \)):

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

Since both limits equal \( f(0) \):

\[ \lim_{x \to 0} f(x) = f(0) = 2 \]

**Conclusion for Statement 3**: The function is continuous at \( x = 0 \), hence this statement is **correct**.

### Step 3: Check differentiability at \( x = 0 \)

To check differentiability at \( x = 0 \), we need to find the left-hand and right-hand derivatives.

**Left-hand derivative**:

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

**Right-hand derivative**:

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

Since the left-hand derivative \(-1\) is not equal to the right-hand derivative \(1\), the function is not differentiable at \( x = 0 \).

**Conclusion for Statement 2**: The function is not differentiable at \( x = 0\), hence this statement is **incorrect**.

### Final Conclusion

- Statement 1: Incorrect
- Statement 2: Incorrect
- Statement 3: Correct

Thus, the only correct statement is **3**.