If `f(x) = {(1+x^(2),",", "when"0 le x lt 1),(1-x,",","when"x gt 1):}`, then

Options

A. `lim_(x to 1^(+)) f(x) != 0`
B. `lim_(x to 1^(+)) f(x) != 2`
C. f(x) is discontinuous at` x = 1`
D. f(x) is continuous at` x = 1`

Answer

To determine whether the function \( f(x) \) is continuous at \( x = 1 \), we need to evaluate the left-hand limit, right-hand limit, and the value of the function at that point.

Given the piecewise function:
\[ f(x) = \begin{cases} 1 + x^2 & \text{if } 0 \leq x < 1 \\ 1 - x & \text{if } x > 1 \end{cases} \]

### Step 1: Calculate \( f(1^-) \)
To find the left-hand limit as \( x \) approaches 1, we use the first piece of the function since it applies when \( x < 1 \):
\[ f(1^-) = \lim_{x \to 1^-} f(x) = 1 + (1)^2 = 1 + 1 = 2 \]

### Step 2: Calculate \( f(1^+) \)
Next, we find the right-hand limit as \( x \) approaches 1, using the second piece of the function since it applies when \( x > 1 \):
\[ f(1^+) = \lim_{x \to 1^+} f(x) = 1 - (1) = 1 - 1 = 0 \]

### Step 3: Evaluate \( f(1) \)
Now we check the value of the function at \( x = 1 \):
\[ f(1) \text{ is not defined since there is no piece of the function for } x = 1. \]

### Step 4: Check for continuity
For \( f(x) \) to be continuous at \( x = 1 \), the following condition must hold:
\[ f(1^-) = f(1^+) = f(1) \]
From our calculations:
- \( f(1^-) = 2 \)
- \( f(1^+) = 0 \)
- \( f(1) \) is not defined.

Since \( f(1^-) \neq f(1^+) \) and \( f(1) \) is not defined, we conclude that the function is discontinuous at \( x = 1 \).

### Final Conclusion
Thus, the function \( f(x) \) is discontinuous at \( x = 1 \).

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