Discuss the continuity of the function :
`f(x)={{:(x",if "xge0),(x^(2)",if "xlt0):}`

Answer

To discuss the continuity of the function

\[ f(x) = \begin{cases} x & \text{if } x \geq 0 \\ x^2 & \text{if } x < 0 \end{cases} \]

we need to check its continuity at the point \(x = 0\). A function is continuous at a point if the following three conditions are satisfied:

1. \(f(a)\) is defined.
2. \(\lim_{x \to a} f(x)\) exists.
3. \(\lim_{x \to a} f(x) = f(a)\).

In this case, we will check these conditions at \(a = 0\).

### Step 1: Evaluate \(f(0)\)

Since \(0\) is greater than or equal to \(0\), we use the first case of the function:

\[ f(0) = 0 \]

**Hint**: Check which piece of the function applies at \(x = 0\).

### Step 2: Calculate the Left-Hand Limit \(\lim_{x \to 0^-} f(x)\)

For the left-hand limit, we consider values of \(x\) that are slightly less than \(0\). In this case, we use the second case of the function:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x^2 \]

As \(x\) approaches \(0\) from the left:

\[ \lim_{x \to 0^-} x^2 = 0^2 = 0 \]

**Hint**: Remember that for \(x < 0\), the function is defined as \(x^2\).

### Step 3: Calculate the Right-Hand Limit \(\lim_{x \to 0^+} f(x)\)

For the right-hand limit, we consider values of \(x\) that are slightly greater than \(0\). In this case, we use the first case of the function:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x \]

As \(x\) approaches \(0\) from the right:

\[ \lim_{x \to 0^+} x = 0 \]

**Hint**: For \(x \geq 0\), the function is defined as \(x\).

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

Now we need to check if the left-hand limit, right-hand limit, and the value of the function at \(0\) are all equal:

- \(f(0) = 0\)
- \(\lim_{x \to 0^-} f(x) = 0\)
- \(\lim_{x \to 0^+} f(x) = 0\)

Since all three values are equal:

\[ f(0) = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = 0 \]

### Conclusion

Thus, the function \(f(x)\) is continuous at \(x = 0\).

**Final Answer**: The function \(f(x)\) is continuous at \(x = 0\).