4.) Consider
\[
f(x)=\left\{\begin{array}{ll}
e^{x} & \text { if } x< 0 \\
x^{2} & \text { if } x \geq 0
\end{array}\right.
\]
Is $f$ continuous at $x=0$ ? Give reasons.

Step 1 ：First, we need to understand the definition of continuity at a point. A function \(f(x)\) is said to be continuous at \(x = a\) if the following three conditions are met: \(f(a)\) is defined, \(\lim_{x\to a^-} f(x)\) exists, \(\lim_{x\to a^+} f(x)\) exists, and \(\lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = f(a)\).

Step 2 ：We are asked to determine if the function \(f(x)\) is continuous at \(x = 0\). We can do this by checking if the above conditions are met at \(x = 0\).

Step 3 ：First, we check if \(f(0)\) is defined. For \(x \geq 0\), \(f(x) = x^2\). So, \(f(0) = (0)^2 = 0\). Therefore, \(f(0)\) is defined.

Step 4 ：Next, we find the left-hand limit, \(\lim_{x\to 0^-} f(x)\). For \(x < 0\), \(f(x) = e^x\). So, \(\lim_{x\to 0^-} f(x) = \lim_{x\to 0^-} e^x = e^0 = 1\).

Step 5 ：Then, we find the right-hand limit, \(\lim_{x\to 0^+} f(x)\). For \(x \geq 0\), \(f(x) = x^2\). So, \(\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} x^2 = (0)^2 = 0\).

Step 6 ：We can see that \(\lim_{x\to 0^-} f(x) \neq \lim_{x\to 0^+} f(x)\), so the function \(f(x)\) is not continuous at \(x = 0\).

Step 7 ：Therefore, the function \(f(x)\) is not continuous at \(x = 0\).

Step 8 ：We can check our result by graphing the function \(f(x)\). The graph will show a jump at \(x = 0\), confirming that \(f(x)\) is not continuous at \(x = 0\).

Step 9 ：So, the final answer is that the function \(f(x)\) is not continuous at \(x = 0\).

Step 10 ：\boxed{\text{The function } f(x) \text{ is not continuous at } x = 0.}