Let` f(x)={(x^2,x in Z),((k(x^2-4))/(2-x),x notinZ):}` Then, `lim_(xrarr2) f(x)`

Options

A. exists only when `k=1`
B. exists for every real k
C. exists for every real k expect `k=1`
D. does not exists

Answer

To solve the limit \( \lim_{x \to 2} f(x) \) where

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

we will focus on the case when \( x \notin \mathbb{Z} \) since we are interested in the limit as \( x \) approaches 2, which is not an integer.

### Step 1: Substitute \( x = 2 \) into the function for \( x \notin \mathbb{Z} \)

We need to evaluate the limit of \( \frac{k(x^2 - 4)}{2 - x} \) as \( x \) approaches 2.

### Step 2: Factor the numerator

The expression \( x^2 - 4 \) can be factored as:

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

Thus, we can rewrite the function:

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

### Step 3: Rewrite the denominator

Notice that \( 2 - x = -(x - 2) \). Therefore, we can rewrite \( f(x) \):

\[ f(x) = \frac{k((x - 2)(x + 2))}{-(x - 2)} = -k(x + 2) \quad \text{for } x \notin \mathbb{Z} \text{ and } x \neq 2 \]

### Step 4: Cancel the common terms

As long as \( x \neq 2 \), we can cancel \( (x - 2) \):

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

### Step 5: Take the limit as \( x \to 2 \)

Now we can find the limit:

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

### Step 6: Conclusion

Thus, the limit is:

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