Answer
$y=\frac{1}{x-2}-3x$ is continous as $x\in(-\infty,2)\cup(2,\infty)$
Work Step by Step
$$y=\frac{1}{x-2}-3x$$
- Domain: $(-\infty,2)\cup(2,\infty)$
We would find all the points where the function is discontinous first, then all the remaining points would be where the function is continous.
We see that at $x=2$: $$\lim_{x\to2}y=\lim_{x\to2}\Big(\frac{1}{x-2}-3x\Big)$$
As $x\to2$, $(x-2)\to0$, meaning that $\frac{1}{x-2}$ will approach infinity, not any single value.
In other words, $\lim_{x\to2}\Big(\frac{1}{x-2}-3x\Big)$ does not exist.
So the function is discontinous at $x=2$. That is also the only point where $y$ is discontinous.
Therefore, $y=\frac{1}{x-2}-3x$ is continous in its domain, $(-\infty,2)\cup(2,\infty)$
