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

Options

A. `5/2`
B. `5`
C. `2`
D. `1/2`

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 (LHL) and the right-hand limit (RHL) at \( x = 2 \) are equal.

The function \( f(x) \) is defined as follows:
- \( f(x) = 2x + 3 \) for \( x \leq 2 \)
- \( f(x) = x + 2k \) for \( x > 2 \)

### Step 1: Calculate the Left-Hand Limit (LHL)

The left-hand limit as \( x \) approaches 2 from the left (i.e., \( x \to 2^- \)) is given by the expression for \( f(x) \) when \( x \leq 2 \):

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2} (2x + 3) \]

Substituting \( x = 2 \):

\[ = 2(2) + 3 = 4 + 3 = 7 \]

### Step 2: Calculate the Right-Hand Limit (RHL)

The right-hand limit as \( x \) approaches 2 from the right (i.e., \( x \to 2^+ \)) is given by the expression for \( f(x) \) when \( x > 2 \):

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

Substituting \( x = 2 \):

\[ = 2 + 2k \]

### Step 3: Set the LHL equal to the RHL

For the limit to exist, the left-hand limit must equal the right-hand limit:

\[ 7 = 2 + 2k \]

### Step 4: Solve for \( k \)

Now, we solve the equation for \( k \):

\[ 7 - 2 = 2k \]

\[ 5 = 2k \]

\[ k = \frac{5}{2} \]

### Final Answer

Thus, the value of \( k \) is:

\[ \boxed{\frac{5}{2}} \]