For what value of k, the function
`f(x) ={:{((x^2-4)/(x-2)", " x ne 2),(" "k", " x=2):},`
is continuous at x =2.

Answer

To determine the value of \( k \) such that the function

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

is continuous at \( x = 2 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 2 equals \( f(2) \).

### Step 1: Find the limit of \( f(x) \) as \( x \) approaches 2.

For \( x \neq 2 \), we can simplify \( f(x) \):

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

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

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

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

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

For \( x \neq 2 \), the \( (x - 2) \) terms cancel out:

\[ f(x) = x + 2 \quad \text{for } x \neq 2 \]

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

Now we find the limit:

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

### Step 3: Set the limit equal to \( f(2) \).

For the function to be continuous at \( x = 2 \), we need:

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

Substituting the values we found:

\[ 4 = k \]

### Conclusion

Thus, the value of \( k \) that makes the function continuous at \( x = 2 \) is:

\[ \boxed{4} \]