if `f(x)={{:(2+x,xge0),(2-x,xlt0):}` then `lim_(x to 0) f(f(x))` is equal to (A) `0` (B) `4` (C) `2` (D) does not exist

Options

A. 0
B. 4
C. 2
D. does not exist

Explanation

To solve the problem, we need to find the limit of \( f(f(x)) \) as \( x \) approaches \( 0 \). The function \( f(x) \) is defined as follows:

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

### Step 1: Determine \( f(f(x)) \)

We will evaluate \( f(f(x)) \) for both cases: when \( x \geq 0 \) and when \( x < 0 \).

#### Case 1: \( x \geq 0 \)

If \( x \geq 0 \), then:

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

Now, we need to find \( f(f(x)) \):

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

Since \( 2 + x \) is also greater than or equal to \( 0 \) (because \( x \geq 0 \)), we use the first case of the function:

\[ f(2 + x) = 2 + (2 + x) = 4 + x \]

So, for \( x \geq 0 \):

\[ f(f(x)) = 4 + x \]

#### Case 2: \( x < 0 \)

If \( x < 0 \), then:

\[ f(x) = 2 - x \]

Now, we need to find \( f(f(x)) \):

\[ f(f(x)) = f(2 - x) \]

Since \( 2 - x \) is greater than \( 0 \) (because \( x < 0 \)), we again use the first case of the function:

\[ f(2 - x) = 2 + (2 - x) = 4 - x \]

So, for \( x < 0 \):

\[ f(f(x)) = 4 - x \]

### Step 2: Find the limit as \( x \) approaches \( 0 \)

Now we need to find \( \lim_{x \to 0} f(f(x)) \).

#### Right-hand limit (as \( x \to 0^+ \))

For \( x \geq 0 \):

\[ f(f(x)) = 4 + x \]

Taking the limit as \( x \) approaches \( 0 \) from the right:

\[ \lim_{x \to 0^+} (4 + x) = 4 + 0 = 4 \]

#### Left-hand limit (as \( x \to 0^- \))

For \( x < 0 \):

\[ f(f(x)) = 4 - x \]

Taking the limit as \( x \) approaches \( 0 \) from the left:

\[ \lim_{x \to 0^-} (4 - x) = 4 - 0 = 4 \]

### Step 3: Conclusion

Both the right-hand limit and left-hand limit as \( x \) approaches \( 0 \) are equal to \( 4 \):

\[ \lim_{x \to 0} f(f(x)) = 4 \]

Thus, the answer is:

**(B) 4**