Examine the continuity of the following function at the indicated pionts.
`f(x)={{:(,x-[x] x ne 1),(,0 x =1):}" at x =0"`

Answer

To examine the continuity of the function \( f(x) = x - [x] \) for \( x \neq 1 \) and \( f(1) = 0 \) at the point \( x = 0 \), we will follow these steps:

### Step 1: Understand the function
The function \( f(x) = x - [x] \) represents the fractional part of \( x \), where \( [x] \) is the greatest integer less than or equal to \( x \). For \( x \) in the interval \( [0, 1) \), \( [x] = 0 \), so \( f(x) = x - 0 = x \).

### Step 2: Calculate \( f(0) \)
To check continuity at \( x = 0 \), we first find the value of the function at this point:
\[ f(0) = 0 - [0] = 0 - 0 = 0. \]

### Step 3: Calculate the left-hand limit as \( x \) approaches 0
Next, we calculate the left-hand limit:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x - [x]). \]
For \( x \) approaching 0 from the left, \( [x] = -1 \) (since it is the greatest integer less than \( x \)). Therefore:
\[ f(x) = x - (-1) = x + 1. \]
Thus, the left-hand limit is:
\[ \lim_{x \to 0^-} f(x) = 0 + 1 = 1. \]

### Step 4: Calculate the right-hand limit as \( x \) approaches 0
Now, we calculate the right-hand limit:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x - [x]). \]
For \( x \) approaching 0 from the right, \( [x] = 0 \). Therefore:
\[ f(x) = x - 0 = x. \]
Thus, the right-hand limit is:
\[ \lim_{x \to 0^+} f(x) = 0. \]

### Step 5: Compare the limits and the function value
Now we compare the left-hand limit, right-hand limit, and the function value at \( x = 0 \):
- Left-hand limit: \( \lim_{x \to 0^-} f(x) = 1 \)
- Right-hand limit: \( \lim_{x \to 0^+} f(x) = 0 \)
- Function value: \( f(0) = 0 \)

### Conclusion
Since the left-hand limit and the right-hand limit are not equal (\( 1 \neq 0 \)), the function \( f(x) \) is not continuous at \( x = 0 \).