Discuss the continuity of the function :
`f(x)={{:((|x-2|)/(x-2)", "xne2),(1" ,"x=2):}` at x = 2

Answer

To determine the continuity of the function

\[ f(x) = \begin{cases} \frac{|x-2|}{x-2} & \text{if } x \neq 2 \\ 1 & \text{if } x = 2 \end{cases} \]

at \( x = 2 \), we need to check the following three conditions:

1. \( f(2) \) is defined.
2. The limit \( \lim_{x \to 2} f(x) \) exists.
3. \( \lim_{x \to 2} f(x) = f(2) \).

### Step 1: Check if \( f(2) \) is defined

From the function definition, we have:

\[ f(2) = 1 \]

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

We need to evaluate the limit from both sides (left-hand limit and right-hand limit).

#### Left-hand limit (\( x \to 2^- \))

When \( x < 2 \), \( |x-2| = -(x-2) = 2-x \). Therefore,

\[ f(x) = \frac{|x-2|}{x-2} = \frac{2-x}{x-2} = \frac{-(x-2)}{x-2} = -1 \]

Thus,

\[ \lim_{x \to 2^-} f(x) = -1 \]

#### Right-hand limit (\( x \to 2^+ \))

When \( x > 2 \), \( |x-2| = x-2 \). Therefore,

\[ f(x) = \frac{|x-2|}{x-2} = \frac{x-2}{x-2} = 1 \]

Thus,

\[ \lim_{x \to 2^+} f(x) = 1 \]

### Step 3: Check if the limit exists

Since the left-hand limit and right-hand limit are not equal:

\[ \lim_{x \to 2^-} f(x) = -1 \quad \text{and} \quad \lim_{x \to 2^+} f(x) = 1 \]

The overall limit \( \lim_{x \to 2} f(x) \) does not exist.

### Conclusion

Since the limit does not exist, we conclude that the function \( f(x) \) is not continuous at \( x = 2 \).

### Summary of Steps:
1. Check if \( f(2) \) is defined: \( f(2) = 1 \).
2. Find left-hand limit: \( \lim_{x \to 2^-} f(x) = -1 \).
3. Find right-hand limit: \( \lim_{x \to 2^+} f(x) = 1 \).
4. Since the limits do not match, conclude that \( f(x) \) is not continuous at \( x = 2 \).