Answer
$ f(x)$ is continuous on
$$(-\infty,-2)\cup(-2,\infty)$$
Work Step by Step
Given $$
f(x)=\left\{\begin{array}{ll}{\frac{x^{3}-8}{x^2-4},} & {x \neq 2, x\neq-2} \\ {3,} & {x=2}\\ {4,} & {x=-2}\end{array}\right.
$$
A function $ f(x)$ is continuous at a point $ x=c $ if and only if it meets the following three conditions:
1. $ f(x)$ exists
2. $\lim\limits _{x \rightarrow c} f(x)$ exists
3. $\lim \limits_{x \rightarrow c} f(x)=f(c)$
Firstly, at $ x=2$
\begin{aligned}a)\lim\limits _{x \rightarrow 2} f(x) &= \lim\limits _{x \rightarrow 2}\frac{x^{3}-8}{x^2-4}\\
&=\lim\limits _{x \rightarrow 2} \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}\\
&=\lim\limits _{x \rightarrow 2} \frac{x^2+2x+4}{ x+2}\\
&= \frac{2^2+4+4}{ 2+2}\\
&=\frac{12}{4}\\
&=3\\
\end{aligned}
$$ b) f(2)=3$$
From (a), (b) since $\lim \limits_{x \rightarrow 3} f(x)=f(2)=3,$ the function is continuous at $ x=2$.
Secondly, at $ x=-2$
\begin{aligned}c)\lim\limits _{x \rightarrow -2} f(x) &= \lim\limits _{x \rightarrow -2}\frac{x^{3}-8}{x^2-4}\\
&=\lim\limits _{x \rightarrow -2} \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}\\
&=\lim\limits _{x \rightarrow -2} \frac{x^2+2x+4}{ x+2}\\
&= \frac{-2^2-4+4}{ -2+2}\\
&=\infty \end{aligned}
$$ d)f(-2)=4$$
From (c), (d) since $\infty=\lim \limits_{x \rightarrow -2} f(x)\neq f(-2)=4,$ the function is not continuous at $ x=-2$.
Thus $ f(x)$ is continuous on
$$(-\infty,-2)\cup(-2,\infty)$$
