Answer
$f(x)$ is continuous on $(-\infty,-2)\cup(-2,\infty)$
Work Step by Step
$f(x)=\frac{x^3-8}{x^2-4}$ for $x\ne2$ and $x\ne-2$
$f(x)=3$ for $x=2$
$f(x)=4$ for $x=-2$
- Domain: $f(x)$ is defined on $(-\infty,\infty)$
1) Is $f(x)$ continuous at $x=2$ and $x=-2$?
- $f(x)$ is continuous at $x=2$ if and only if $\lim_{x\to2}f(x)=f(2)=3$. Similarly, $f(x)$ is continuous at $x=-2$ if and only if $\lim_{x\to-2}f(x)=f(-2)=4$
- First, we would examine the function of $f(x)=\frac{x^3-8}{x^2-4}$:
$$f(x)=\frac{x^3-8}{x^2-4}=\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}=\frac{x^2+2x+4}{x+2}$$
Therefore,
$$\lim_{x\to2}f(x)=\lim_{x\to2}\frac{x^2+2x+4}{x+2}$$
$$\lim_{x\to2}f(x)=\frac{2^2+2\times2+4}{2+2}=\frac{12}{4}=3$$
So $f(x)$ is continuous at $x=2$.
$$\lim_{x\to-2}f(x)=\lim_{x\to-2}\frac{x^2+2x+4}{x+2}$$
As $x\to-2$, $(x+2)$ approaches $0$, which means $\frac{x^2+2x+4}{x+2}$, instead of approaching a definite value, would go for $\infty$.
So $\lim_{x\to-2}f(x)$ does not exist. $f(x)$, hence, is not continuous at $x=-2$.
2) Is $f(x)$ continuous at other points?
As $x\ne2$ and $x\ne-2$:
- $\lim_{x\to c}(x^3-8)=c^3-8$
- $\lim_{x\to c}(x^2-4)=c^2-4$
So both $y=x^3-8$ and $y=x^2-4$ are continuous on $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$
Applying Theorem 8, the quotient of 2 functions continuous at $x=c$ is also continuous at $x=c$
Therefore, $f(x)=\frac{x^3-8}{x^2-4}$ is continuous on $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$.
In conclusion, combining the results from 1) and 2), $f(x)$ is continuous on $(-\infty,-2)\cup(-2,\infty)$
