Is $|g(n)| \le c \cdot |f(n)| \iff \lim_{n \to \infty} \frac {g(n)} {f(n)} = c $ true? I want to show that:
$|g(n)| \le c \cdot |f(n)| \iff \lim_{n \to \infty} \frac {g(n)} {f(n)} = c\mid c \in \mathbb R_0^+ \land n \in \mathbb N \land f,g: \mathbb N \to \mathbb R_0^+$ 
Due to the defined sets it's the same as:
$g(n) \le c \cdot f(n) \iff \lim_{n \to \infty} \frac{g(n)}{f(n)} = c$
Now I can make it look similar:
$g(n) \le c \cdot f(n) \iff \lim_{n \to \infty} g(n) = \lim_{n \to \infty}  c \cdot f(n)$
But I'm not feeling confident enough about the rules of limits to determine the given term as true.
 A: Example: $g(n)=n$ and $f(n)=3n$. Then $g(n) \le f(n)$  ($c=1$)
But $\frac{g(n)}{f(n)}= 1/3$
A: No, that is not correct. First, the left hand side has to hold only asymptotically, i.e., for all $n \geq n_0$.
Definition (Limit). We say that a real sequence $(x(n))_n$ converges to $c$ if for every $\epsilon>0$ there is a $n_0 = n_0(\epsilon)$ so that $|x(n) - c| < \epsilon$ for all $n\geq 0$. (Source: Wikipedia)
Then, from the definition of the limit, since $g(n)/f(n) \to c$, for every $\epsilon$ there is a $n_0$ so that for all $n\geq n_0$, $|g(n)/f(n) - c|<\epsilon$, which implies that
$$
\begin{aligned}
&\left|\frac{g(n)}{f(n)} - c\right|<\epsilon\\
\Rightarrow&\frac{g(n)}{f(n)} - c\leq  \epsilon.\end{aligned}
$$
And taking the infimum with respect to $\epsilon>0$ on both sides we have
$$
\frac{g(n)}{f(n)} \leq c\tag{1}\label{1}
$$
Now, for the converse, if you assume that \eqref{1} holds, you only have
$$
\limsup_{n\to\infty}\frac{g(n)}{f(n)} \leq c\tag{2}\label{2},
$$
and if the limit exists, $\lim_{n\to\infty}{g(n)}/{f(n)} \leq c$
A: A major flaw in your assertion I feel have not been touched upon yet: the uniqueness of $c$
The first statement says that $|g(n)| \leq c |f(n)|$. Such $c$ is never unique, due to the following inequalities:
$$|g(n)| \leq c|f(n)| \leq c|f(n)| + c'|f(n)| = (c+c')|f(n)|$$
where $c' \geq 0$. If the inequality holds for some $c$, it also holds for all $c' \in [c,+\infty)$. 
Howver, the second statement is $$\lim_{n \to \infty} \frac{g(n)}{f(n)} = c.$$
But a limit, if it exists, is unique! It cannot be all values in the interval $[c,+\infty)$ at the same time!
A: (1).If $c>0$ then $|g(n)|\leq c|f(n)|$ does not imply that $\lim_{n\to \infty}|g(n)/f(n)|$ exists. E.g. $f(n)=2$ when $n$ is even, and $f(n)=1$ when $n$ is odd, with $g(n)=1$ for all $n,$ and $c=2.$ 
(2). If $c>0$ and $|g(n)|<c|f(n)|$ and if the limit $\lim_{n\to \infty}|g(n)/f(n)|$ does exist, the limit may be any $r\in[0,c].$ E.g. for $r\in (0,c]$ let $g(n)=rf(n)$ for all $n.$ And for $r=0$ let $g(n)=cf(n)/n$ for $n\in \mathbb N.$
