Show that the function $f$ given by $f(x) = \begin{cases} x^3 + 3, & \text{if } x \neq 0 \\ 1, & \text{if } x = 0 \end{cases}$ is not continuous at $x = 0$.

(N/A) function $f(x)$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$.
$1$. First,we find the value of the function at $x = 0$:
$f(0) = 1$.
$2$. Next,we find the limit of the function as $x$ approaches $0$:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^3 + 3) = 0^3 + 3 = 3$.
$3$. Comparing the limit and the function value:
Since $\lim_{x \to 0} f(x) = 3$ and $f(0) = 1$,we see that $\lim_{x \to 0} f(x) \neq f(0)$.
Therefore,the function $f$ is not continuous at $x = 0$.