<s>[INST] Domain and range of $g(x) = \frac{3x^2+4}{x-2}$ 
$$g(x) = \frac{3x^2+4}{x-2}$$

for the domain, I find, $\{x|x\ne 2\}$'
and for the range
$y = \dfrac{3x^2+4}{x-2}$ and when $x=2-\epsilon$, $y \to \dfrac{3(2-\epsilon)^2 + 4}{(2-\epsilon)-2} = \dfrac{K}{-\epsilon} \to -\infty$
When $x = 2+\epsilon$, $y \to \dfrac{3(2+\epsilon)^2+4}{(2+\epsilon)-2} = \dfrac{K}{\epsilon} \to \infty$
Therefore, $\text{ran}(g(x)) = (-\infty, \infty)$ 
have I done this right or am I missing something? For instance, I thought about the case when $x \to \infty$, but this looks like $y \to 1$ in this case, nothing remarkable. I don't even know if that is what we say when we have an $\frac{\infty^2}{\infty}$....Do we just say that is one? Or do we say that is $\infty$ still?
 [/INST] Fix $y \in \mathbb{R}$. Then you look for $x$ such that
$$\frac{3x^2+4}{x-2}=y.$$
Solve this equation: you get
$$3x^2+4=xy-2y \Rightarrow 3x^2-yx+2y+4=0,$$
which gives 
$$x=\frac{y\pm\sqrt{y^2-24y-48}}{6}. $$
So your equation admits real solutions only if $y^2-24y-48 \geq 0$. But this is $y^2-24y+144-192 = (y-12)^2-192 \geq 0$, i.e. $\lvert y-12 \rvert \geq \sqrt{192}$. It is clear that this is not true for every $y$, since for example $y=0$ does not satisfy the inequality. So the range is not $\mathbb{R}$, but is given by those points $y$ satisfying $\lvert y-12 \rvert \geq \sqrt{192}$.
One more comment on your last observation. If $x \to \infty$ then $$g(x) = \frac{x^2\left(3+\frac{4}{x^2}\right)}{x\left(1-\frac{2}{x}\right)} = \frac{x\left(3+\frac{4}{x^2}\right)}{\left(1-\frac{2}{x}\right)} \to \frac{\infty \cdot 3}{1}=\infty.$$
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