Let` f(x)={(x^2,x notin Z),((k(x^2-4))/(2-x),x notinZ):}`
Then, `lim_(xto2) f(x)`

Options

A. exists only when `k=1`
B. exists for every real k
C. exists for every real k except `k=1`
D. Does not exist

Answer

To solve the problem, we need to find the limit of the function \( f(x) \) as \( x \) approaches 2. The function is defined as follows:

\[ f(x) = \begin{cases} x^2 & \text{if } x \notin \mathbb{Z} \\ \frac{k(x^2 - 4)}{2 - x} & \text{if } x \in \mathbb{Z} \end{cases} \]

Since we are interested in the limit as \( x \) approaches 2, we will consider the case where \( x \) is not an integer (as \( x \) approaches 2, it will be very close to 2 but not equal to 2).

### Step 1: Analyze the function for \( x \notin \mathbb{Z} \)

For \( x \notin \mathbb{Z} \), we have:

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

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

We can directly substitute \( x = 2 \) into the function \( f(x) \):

\[ \lim_{x \to 2} f(x) = \lim_{x \to 2} x^2 = 2^2 = 4 \]

### Step 3: Analyze the function for \( x \in \mathbb{Z} \)

Now, we also need to consider the case when \( x \) is an integer. Specifically, we will look at the limit of the second part of the function as \( x \) approaches 2.

\[ f(x) = \frac{k(x^2 - 4)}{2 - x} \]

### Step 4: Simplify the expression

We recognize that \( x^2 - 4 \) can be factored:

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

Thus, we can rewrite \( f(x) \) as:

\[ f(x) = \frac{k((x - 2)(x + 2))}{2 - x} \]

### Step 5: Factor out the common terms

Notice that \( 2 - x = -(x - 2) \), so we can rewrite the function as:

\[ f(x) = \frac{k(x - 2)(x + 2)}{-(x - 2)} = -k(x + 2) \]

### Step 6: Take the limit as \( x \) approaches 2

Now, we can find the limit:

\[ \lim_{x \to 2} f(x) = \lim_{x \to 2} -k(x + 2) = -k(2 + 2) = -4k \]

### Step 7: Conclusion

For the limit \( \lim_{x \to 2} f(x) \) to exist, it must be equal from both cases (when \( x \) is an integer and when \( x \) is not an integer):

\[ 4 = -4k \]

Solving for \( k \):

\[ k = -1 \]

### Final Result

Thus, the limit exists for every real value of \( k \), but specifically, we find that when \( k = -1 \), both cases yield the same limit.

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