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 $x=0$.

Step 1: Recall the definition of continuity

For a function to be continuous at a point $x=a$, three conditions must be met: the function must be defined at $a$, the limit of the function as $x$ approaches $a$ must exist, and this limit must be equal to the function's value at $a$. In this problem, we need to check these conditions at $x=0$.

Step 2: Find the function value at $x=0$

From the given definition of the function, when $x=0$, the function $f(x)$ is defined as $1$. So, $f(0)=1$.

Step 3: Find the limit of the function as x approaches 0

To find the limit as $x$ approaches $0$, we consider the part of the function definition for $x \neq 0$, which is $f(x) = x^3 + 3$. We substitute $x=0$ into this expression to evaluate the limit.

Step 4: Evaluate the limit

Substituting $x=0$ into the expression $x^3 + 3$, we get $0^3 + 3 = 0 + 3 = 3$. Therefore, the limit of $f(x)$ as $x$ approaches $0$ is $3$.

Step 5: Compare the limit with the function value

We found that the limit of $f(x)$ as $x$ approaches $0$ is $3$, but the function value at $x=0$ is $1$. Since these two values are not equal, the condition for continuity is not met.

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