Discuss the continuity of the function :
`f(x)={{:(x", "0lexlt1/2),(1/2", "x=1/2),(1-x", "1/2ltxle1):}` at `x=1/2`

Answer

To determine the continuity of the function

\[ f(x) = \begin{cases} x & \text{for } 0 \leq x < \frac{1}{2} \\ \frac{1}{2} & \text{for } x = \frac{1}{2} \\ 1 - x & \text{for } \frac{1}{2} < x \leq 1 \end{cases} \]

at \( x = \frac{1}{2} \), we need to check the following conditions:

1. The function \( f(x) \) must be defined at \( x = \frac{1}{2} \).
2. The left-hand limit as \( x \) approaches \( \frac{1}{2} \) must exist.
3. The right-hand limit as \( x \) approaches \( \frac{1}{2} \) must exist.
4. The left-hand limit, right-hand limit, and the value of the function at \( x = \frac{1}{2} \) must all be equal.

### Step 1: Check if \( f\left(\frac{1}{2}\right) \) is defined

From the function definition, we see that:

\[ f\left(\frac{1}{2}\right) = \frac{1}{2} \]

### Step 2: Calculate the left-hand limit as \( x \) approaches \( \frac{1}{2} \)

The left-hand limit is given by:

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

### Step 3: Calculate the right-hand limit as \( x \) approaches \( \frac{1}{2} \)

The right-hand limit is given by:

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

### Step 4: Compare the limits and the function value

Now we have:

- Left-hand limit: \( \lim_{x \to \frac{1}{2}^- f(x) = \frac{1}{2} \)
- Right-hand limit: \( \lim_{x \to \frac{1}{2}^+ f(x) = \frac{1}{2} \)
- Function value: \( f\left(\frac{1}{2}\right) = \frac{1}{2} \)

Since all three values are equal:

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

### Conclusion

The function \( f(x) \) is continuous at \( x = \frac{1}{2} \).

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