This is just a minor part of another problem and intuitively clear, but I think that it is necessary to have a proof of this assertion.
If $\lim_{x\to \infty} f(x) = c\neq 0$ and $\lim_{x\to \infty} g(x) = 0$, then $\lim_{x\to \infty} \frac{f(x)}{g(x)}$ does not exist.
Proof:
$(\lim_{x\to \infty} f(x) = c\neq 0)\Rightarrow (\lim_{x\to \infty} \frac{1}{f(x)}=\frac{1}{c})$; this is already proven. $(\lim_{x\to \infty} \frac{1}{f(x)}=\frac{1}{c})\land(\lim_{x\to \infty} g(x) = 0)\Rightarrow (\lim_{x\to \infty} \frac{g(x)}{f(x)}=\frac{1}{c}\cdot0=0)$; this is also already proven. Now, $\lim_{x\to \infty} \frac{g(x)}{f(x)}=0$ means that for every $\varepsilon> 0$ there is a $N\in\mathbb{R}_{>0}$ such that, for all x, $$\textrm{if } x>N,\textrm{then}\quad\left|\frac{g(x)}{f(x)}-0\right|=\left|\frac{g(x)}{f(x)}\right|=\frac{|g(x)|}{|f(x)|}<\epsilon, $$$$\textrm{so } \frac{1}{\epsilon}< \frac{|f(x)|}{|g(x)|}=\left|\frac{f(x)}{g(x)}\right|. $$ Thus, $\lim_{x\to \infty} \left|\frac{f(x)}{g(x)}\right|=\infty$.
Since $\lim_{x\to \infty} h(x) = 0$ if and only if $\lim_{x\to \infty} |h(x)| = 0$ (this can be easily proved), $\lim_{x\to \infty} \frac{f(x)}{g(x)}\neq0$. Suppose $\lim_{x\to \infty} \frac{f(x)}{g(x)}=l\neq0$, then $$\lim_{x\to \infty}f(x)=\lim_{x\to \infty} \frac{f(x)}{g(x)} \cdot\lim_{x\to \infty}g(x)=l\cdot0=0,$$ a contradiction! $\square$
First of all is this proof correct? It seems a bit long for this relatively easy assertion, so are there any superfluous parts in it? Would it be sufficient to end with the statement $\lim_{x\to \infty} \left|\frac{f(x)}{g(x)}\right|=\infty$?
An easier proof: Prove by contradiction. (If $g(x)=0$ for $x$ arbitrarily large then $\frac {f(x)} { g(x)}$ won't even exist, so assume that $g(x) \neq 0 $ for large enough $x$). Now $f(x)=\frac {f(x)} {g(x)} g(x)$ So if $l=\lim \frac {f(x)} {g(x)}$ exists and is finite and $g(x) \to 0$ then $f(x) \to (l)(0)=0$ a contradiction.