What 😂😂😂 are you answering

My apologies for the previous incorrect response. I will now answer the question in the image provided.

The question asks to find the value of $k$ for which the function $f(x)$ is continuous at $x=2$.

The function is given by: $f(x) = \begin{cases} \frac{x^{2} - 4}{x - 2} & for x \neq 2 \\ k & for x = 2 \end{cases}$

For $f(x)$ to be continuous at $x=2$, the following condition must be satisfied: $\lim_{x \to 2} f(x) = f(2)$

Step 1: Find $f(2)$. From the definition of the function, when $x=2$, $f(x) = k$. So, $f(2) = k$.

Step 2: Find $\lim_{x \to 2} f(x)$. For $x \neq 2$, $f(x) = \frac{x^{2} - 4}{x - 2}$. We can factor the numerator: $x^{2} - 4 = (x - 2)(x + 2)$. So, for $x \neq 2$: $f(x) = \frac{(x - 2)(x + 2)}{x - 2} = x + 2$ Now, we can find the limit: $\lim_{x \to 2} f(x) = \lim_{x \to 2} (x + 2)$ Substitute $x=2$ into the expression: $\lim_{x \to 2} (x + 2) = 2 + 2 = 4$

Step 3: Equate the limit to $f(2)$ to find $k$. For continuity at $x=2$: $\lim_{x \to 2} f(x) = f(2)$ $4 = k$

The value of $k$ for which the function is continuous at $x=2$ is $\boxed{*4*}$.