Let a function f be defined as
`f(x) = {:{(x," if " 0 le x lt 1/2),(0," if "x=1/2),(x-1," if" 1/2 lt x le 1):}`
Establish the existence of `Lim_(x to 1/2) f(x)` .

Answer

To establish the existence of the limit \( \lim_{x \to \frac{1}{2}} f(x) \) for the given piecewise function

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

we need to find both the left-hand limit and the right-hand limit as \( x \) approaches \( \frac{1}{2} \).

### Step 1: Calculate the Right-Hand Limit

The right-hand limit as \( x \) approaches \( \frac{1}{2} \) is given by:

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

For \( x \) slightly greater than \( \frac{1}{2} \), we use the third case of the function:

\[ f(x) = x - 1 \]

Thus, we can write:

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

Substituting \( x = \frac{1}{2} + h \) where \( h \to 0^+ \):

\[ \lim_{h \to 0^+} \left(\frac{1}{2} + h - 1\right) = \lim_{h \to 0^+} \left(h - \frac{1}{2}\right) = -\frac{1}{2} \]

### Step 2: Calculate the Left-Hand Limit

The left-hand limit as \( x \) approaches \( \frac{1}{2} \) is given by:

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

For \( x \) slightly less than \( \frac{1}{2} \), we use the first case of the function:

\[ f(x) = x \]

Thus, we can write:

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

Substituting \( x = \frac{1}{2} - h \) where \( h \to 0^+ \):

\[ \lim_{h \to 0^+} \left(\frac{1}{2} - h\right) = \frac{1}{2} \]

### Step 3: Compare the Limits

Now we compare the left-hand limit and the right-hand limit:

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

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

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

### Conclusion

Thus, we conclude that:

\[ \lim_{x \to \frac{1}{2}} f(x) \text{ does not exist.} \]