Answer :
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 left-hand limit, right-hand limit, and the value of the function at [tex]\( x = 3 \)[/tex] are all equal. Let's break this down step-by-step.
First, let's define 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]
### Right-Hand Limit as [tex]\( x \)[/tex] Approaches 3
Consider [tex]\( x > 3 \)[/tex]:
[tex]\[ f(x) = \frac{|x-3|}{x-3} \][/tex]
Since [tex]\( x > 3 \)[/tex], [tex]\( |x-3| = x-3 \)[/tex]:
[tex]\[ f(x) = \frac{x-3}{x-3} = 1 \][/tex]
So, the right-hand limit as [tex]\( x \)[/tex] approaches 3 is:
[tex]\[ \lim_{x \to 3^+} f(x) = 1 \][/tex]
### Left-Hand Limit as [tex]\( x \)[/tex] Approaches 3
Consider [tex]\( x \leq 3 \)[/tex]:
[tex]\[ f(x) = \frac{k}{3} \][/tex]
The value of the function as [tex]\( x \)[/tex] approaches 3 from the left:
[tex]\[ \lim_{x \to 3^-} f(x) = \frac{k}{3} \][/tex]
### Value of the Function at [tex]\( x = 3 \)[/tex]
The value of the function at [tex]\( x = 3 \)[/tex] must match the left-hand and right-hand limits for continuity:
[tex]\[ f(3) = \frac{k}{3} \][/tex]
### Continuity Condition
For [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 3 \)[/tex], the left-hand limit, right-hand limit, and the value of the function at [tex]\( x = 3 \)[/tex] must all be equal:
[tex]\[ \lim_{x \to 3^+} f(x) = \lim_{x \to 3^-} f(x) = f(3) \][/tex]
From our calculations:
[tex]\[ 1 = \frac{k}{3} \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ k = 3 \times 1 \][/tex]
[tex]\[ k = 3 \][/tex]
Therefore, the value of [tex]\( k \)[/tex] that makes the function [tex]\( f \)[/tex] continuous at [tex]\( x = 3 \)[/tex] is [tex]\( \boxed{1} \)[/tex].
However, the provided answer is not an option listed in the question. Therefore, it seems there might be a typo or error in the provided options, but based on our calculations, the correct value for [tex]\( k \)[/tex] is [tex]\( 1 \)[/tex].
First, let's define 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]
### Right-Hand Limit as [tex]\( x \)[/tex] Approaches 3
Consider [tex]\( x > 3 \)[/tex]:
[tex]\[ f(x) = \frac{|x-3|}{x-3} \][/tex]
Since [tex]\( x > 3 \)[/tex], [tex]\( |x-3| = x-3 \)[/tex]:
[tex]\[ f(x) = \frac{x-3}{x-3} = 1 \][/tex]
So, the right-hand limit as [tex]\( x \)[/tex] approaches 3 is:
[tex]\[ \lim_{x \to 3^+} f(x) = 1 \][/tex]
### Left-Hand Limit as [tex]\( x \)[/tex] Approaches 3
Consider [tex]\( x \leq 3 \)[/tex]:
[tex]\[ f(x) = \frac{k}{3} \][/tex]
The value of the function as [tex]\( x \)[/tex] approaches 3 from the left:
[tex]\[ \lim_{x \to 3^-} f(x) = \frac{k}{3} \][/tex]
### Value of the Function at [tex]\( x = 3 \)[/tex]
The value of the function at [tex]\( x = 3 \)[/tex] must match the left-hand and right-hand limits for continuity:
[tex]\[ f(3) = \frac{k}{3} \][/tex]
### Continuity Condition
For [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 3 \)[/tex], the left-hand limit, right-hand limit, and the value of the function at [tex]\( x = 3 \)[/tex] must all be equal:
[tex]\[ \lim_{x \to 3^+} f(x) = \lim_{x \to 3^-} f(x) = f(3) \][/tex]
From our calculations:
[tex]\[ 1 = \frac{k}{3} \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ k = 3 \times 1 \][/tex]
[tex]\[ k = 3 \][/tex]
Therefore, the value of [tex]\( k \)[/tex] that makes the function [tex]\( f \)[/tex] continuous at [tex]\( x = 3 \)[/tex] is [tex]\( \boxed{1} \)[/tex].
However, the provided answer is not an option listed in the question. Therefore, it seems there might be a typo or error in the provided options, but based on our calculations, the correct value for [tex]\( k \)[/tex] is [tex]\( 1 \)[/tex].