`f(x)= {{:((|x-4|)/(2(x-4)), if x ne 4),(0,if x = 4):}` at `x = 4`.

Explanation

To determine the continuity of the function \( f(x) \) at \( x = 4 \), we need to analyze the left-hand limit, right-hand limit, and the value of the function at that point.

The function is defined as:
\[ f(x) = \begin{cases} \frac{|x - 4|}{2(x - 4)} & \text{if } x \neq 4 \\ 0 & \text{if } x = 4 \end{cases} \]

### Step 1: Calculate the left-hand limit as \( x \) approaches 4.
We need to find:
\[ \lim_{x \to 4^-} f(x) = \lim_{x \to 4^-} \frac{|x - 4|}{2(x - 4)} \]
Since \( x \) is approaching 4 from the left, \( |x - 4| = 4 - x \). Thus, we can rewrite the limit:
\[ \lim_{x \to 4^-} \frac{4 - x}{2(x - 4)} = \lim_{x \to 4^-} \frac{-(x - 4)}{2(x - 4)} = \lim_{x \to 4^-} \frac{-1}{2} = -\frac{1}{2} \]

### Step 2: Calculate the right-hand limit as \( x \) approaches 4.
Next, we find:
\[ \lim_{x \to 4^+} f(x) = \lim_{x \to 4^+} \frac{|x - 4|}{2(x - 4)} \]
Since \( x \) is approaching 4 from the right, \( |x - 4| = x - 4 \). Thus, we can rewrite the limit:
\[ \lim_{x \to 4^+} \frac{x - 4}{2(x - 4)} = \lim_{x \to 4^+} \frac{1}{2} = \frac{1}{2} \]

### Step 3: Determine the value of the function at \( x = 4 \).
From the definition of the function:
\[ f(4) = 0 \]

### Step 4: Check for continuity.
For \( f(x) \) to be continuous at \( x = 4 \), the following must hold:
\[ \lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) = f(4) \]
We found:
- \( \lim_{x \to 4^-} f(x) = -\frac{1}{2} \)
- \( \lim_{x \to 4^+} f(x) = \frac{1}{2} \)
- \( f(4) = 0 \)

Since the left-hand limit and right-hand limit are not equal, and neither is equal to \( f(4) \), we conclude that \( f(x) \) is discontinuous at \( x = 4 \).

### Final Conclusion:
The function \( f(x) \) is discontinuous at \( x = 4 \).

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