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ScanSolve AI Tutor · Accepted Answer

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} & \text{for } x \neq 2 \\ k & \text{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)

Find $f(2)$ .
From the definition of the function, when $x=2$ , $f(x) = k$ .
So, $f(2) = k$ .
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:
$\lim_{x \to 2} (x + 2) = 2 + 2 = 4$
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 $\boxed{4}$ .

Answer