Consider the following in respect of the function
`f(x)={{:(2+x,xge0),(2-x,xlt0):}`
1. `underset(xrarr1)limf(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 / are correct?

Options

A. Only 1
B. Only 3
C. Both 2 and 3
D. Both 1 and 3

Answer

To analyze the function \( f(x) \) defined as:

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

we will evaluate the three statements given in the question.

### Step 1: Evaluate the first statement
**Statement 1:** \( \lim_{x \to 1} f(x) \) does not exist.

To find \( \lim_{x \to 1} f(x) \):
- Since \( 1 \geq 0 \), we use the first case of the function:
\[ f(1) = 2 + 1 = 3 \]

Since \( f(x) \) is continuous for \( x \geq 0 \), the limit exists and is equal to \( f(1) \). Thus, the limit exists.

**Conclusion:** Statement 1 is **false**.

### Step 2: Evaluate the second statement
**Statement 2:** \( f(x) \) is differentiable at \( x = 0 \).

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

- **Left-hand derivative (LHD)**:
\[ \text{LHD} = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{(2 - h) - 2}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1 \]

- **Right-hand derivative (RHD)**:
\[ \text{RHD} = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{(2 + h) - 2}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1 \]

Since LHD \( \neq \) RHD, \( f(x) \) is not differentiable at \( x = 0 \).

**Conclusion:** Statement 2 is **false**.

### Step 3: Evaluate the third statement
**Statement 3:** \( f(x) \) is continuous at \( x = 0 \).

To check continuity at \( x = 0 \), we need to verify if:
\[ \lim_{x \to 0} f(x) = f(0) \]

- We already found \( f(0) \):
\[ f(0) = 2 + 0 = 2 \]

- **Left-hand limit** as \( x \to 0^- \):
\[ \lim_{x \to 0^-} f(x) = 2 - 0 = 2 \]

- **Right-hand limit** as \( x \to 0^+ \):
\[ \lim_{x \to 0^+} f(x) = 2 + 0 = 2 \]

Since both limits equal \( f(0) \):
\[ \lim_{x \to 0} f(x) = 2 = f(0) \]

Thus, \( f(x) \) is continuous at \( x = 0 \).

**Conclusion:** Statement 3 is **true**.

### Final Conclusion
Only Statement 3 is correct. Therefore, the answer is:

**Option B: Only 3 is correct.**