To determine the value of [tex]\( k \)[/tex] that makes the function [tex]\( f \)[/tex] continuous at [tex]\( x = 3 \)[/tex], we need to ensure that the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 3 from both sides is equal to [tex]\( f(3) \)[/tex].
Given the function:
[tex]\[
f(x) = \begin{cases}
\frac{|x - 3|}{x - 3} & \text{if } x > 3 \\
\frac{k}{3} & \text{if } x \leq 3
\end{cases}
\][/tex]
Let's examine this step-by-step.
### Step 1: Evaluate the right-hand limit ([tex]\( x > 3 \)[/tex])
For [tex]\( x > 3 \)[/tex], the function is defined as [tex]\( f(x) = \frac{|x - 3|}{x - 3} \)[/tex].
When [tex]\( x > 3 \)[/tex], the absolute value [tex]\( |x - 3| = x - 3 \)[/tex]. Thus,
[tex]\[
f(x) = \frac{x - 3}{x - 3} = 1
\][/tex]
So, the limit as [tex]\( x \)[/tex] approaches 3 from the right (denoted as [tex]\( \lim_{x \to 3^+} f(x) \)[/tex]) is:
[tex]\[
\lim_{x \to 3^+} f(x) = 1
\][/tex]
### Step 2: Evaluate the left-hand limit ([tex]\( x \leq 3 \)[/tex])
For [tex]\( x \leq 3 \)[/tex], the function is defined as [tex]\( f(x) = \frac{k}{3} \)[/tex].
This is a constant function, so the expression does not change as [tex]\( x \)[/tex] approaches 3 from the left. Therefore,
[tex]\[
\lim_{x \to 3^-} f(x) = \frac{k}{3}
\][/tex]
### Step 3: Ensure continuity at [tex]\( x = 3 \)[/tex]
For [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 3 \)[/tex], the right-hand limit must equal the left-hand limit and also be equal to [tex]\( f(3) \)[/tex].
Given:
[tex]\[
\lim_{x \to 3^+} f(x) = 1 \quad \text{and} \quad \lim_{x \to 3^-} f(x) = \frac{k}{3}
\][/tex]
We set these equal for continuity:
[tex]\[
1 = \frac{k}{3}
\][/tex]
### Step 4: Solve for [tex]\( k \)[/tex]
To find [tex]\( k \)[/tex], solve the equation:
[tex]\[
1 = \frac{k}{3}
\][/tex]
Multiply both sides by 3:
[tex]\[
k = 3
\][/tex]
### Conclusion
The function [tex]\( f \)[/tex] is continuous at [tex]\( x = 3 \)[/tex] if [tex]\( k = 3 \)[/tex].
Therefore, the correct answer is [tex]\((A) 3\)[/tex].
