Find 'k' so that :
`lim_(x to 2) f(x)` exists, where :
`f(x)={{:(2x+3",", "if "x le 2), (x+k",", "if "x gt 2):}`

Options

A. 3
B. 4
C. 5
D. 6

Answer

To find the value of \( k \) such that the limit \( \lim_{x \to 2} f(x) \) exists, we need to ensure that the left-hand limit and the right-hand limit at \( x = 2 \) are equal.

Given the function:
\[ f(x) = \begin{cases} 2x + 3 & \text{if } x \leq 2 \\ x + k & \text{if } x > 2 \end{cases} \]

### Step 1: Calculate the left-hand limit as \( x \) approaches 2.
For \( x \leq 2 \):
\[ f(x) = 2x + 3 \]
Thus, the left-hand limit is:
\[ \lim_{x \to 2^-} f(x) = 2(2) + 3 = 4 + 3 = 7 \]

### Step 2: Calculate the right-hand limit as \( x \) approaches 2.
For \( x > 2 \):
\[ f(x) = x + k \]
Thus, the right-hand limit is:
\[ \lim_{x \to 2^+} f(x) = 2 + k \]

### Step 3: Set the left-hand limit equal to the right-hand limit.
For the limit to exist at \( x = 2 \), we need:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \]
This gives us the equation:
\[ 7 = 2 + k \]

### Step 4: Solve for \( k \).
Rearranging the equation:
\[ k = 7 - 2 = 5 \]

### Conclusion:
The value of \( k \) is \( 5 \).