Discuss the continuity of the function f(x)`={(1+x^2,x le 1),(1-x, x gt 1):}` at x=1.

Answer

To discuss the continuity of the function \( f(x) = \begin{cases} 1 + x^2 & \text{if } x \leq 1 \\ 1 - x & \text{if } x > 1 \end{cases} \) at \( x = 1 \), we need to check the following:

1. **Value of the function at \( x = 1 \)**:
\[ f(1) = 1 + 1^2 = 1 + 1 = 2 \]

2. **Left-hand limit as \( x \) approaches 1**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (1 + x^2) = 1 + 1^2 = 2 \]

3. **Right-hand limit as \( x \) approaches 1**:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (1 - x) = 1 - 1 = 0 \]

4. **Conclusion**:
For the function to be continuous at \( x = 1 \), the following must hold:
\[ \lim_{x \to 1^-} f(x) = f(1) = \lim_{x \to 1^+} f(x) \]
We found:
- \( \lim_{x \to 1^-} f(x) = 2 \)
- \( f(1) = 2 \)
- \( \lim_{x \to 1^+} f(x) = 0 \)

Since \( \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x) \), the function is discontinuous at \( x = 1 \).

### Summary:
The function \( f(x) \) is discontinuous at \( x = 1 \).

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