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 conditions:

1. The left-hand limit (LHL) as \( x \) approaches 1.
2. The right-hand limit (RHL) as \( x \) approaches 1.
3. The value of the function at \( x = 1 \).

### Step 1: Calculate the Left-Hand Limit (LHL)

The left-hand limit is given by:

\[ \text{LHL} = \lim_{x \to 1^-} f(x) \]

Since we are approaching from the left (where \( x \leq 1 \)), we use the first case of the function:

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

Substituting \( x = 1 \):

\[ \text{LHL} = 1 + 1^2 = 1 + 1 = 2 \]

### Step 2: Calculate the Right-Hand Limit (RHL)

The right-hand limit is given by:

\[ \text{RHL} = \lim_{x \to 1^+} f(x) \]

Since we are approaching from the right (where \( x > 1 \)), we use the second case of the function:

\[ \text{RHL} = \lim_{x \to 1^+} (1 - x) \]

Substituting \( x = 1 \):

\[ \text{RHL} = 1 - 1 = 0 \]

### Step 3: Calculate the Value of the Function at \( x = 1 \)

Now we find the value of the function at \( x = 1 \):

\[ f(1) = 1 + 1^2 = 1 + 1 = 2 \]

### Step 4: Compare the Limits and the Function Value

Now we compare the LHL, RHL, and \( f(1) \):

- LHL = 2
- RHL = 0
- \( f(1) = 2 \)

### Conclusion

Since LHL \( \neq \) RHL (2 \( \neq \) 0), and RHL \( \neq f(1) \) (0 \( \neq \) 2), the function is not continuous at \( x = 1 \).

Thus, we conclude that:

\[ f(x) \text{ is discontinuous at } x = 1. \]

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