Suppose that we know that the following limit is defined: $$\lim_{n\to \infty} \frac{f(n)}{g(n)}=l$$ Where $g,f : \mathbb{N} \to\mathbb{R}$.
And, also, we are aware of the following:
$\exists \: c \in \mathbb{R},\: n_0 \in \mathbb{N}: f(n)\le c\cdot g(n) \ \forall \ n \ge n_0 $
$\exists \: c' \in \mathbb{R},\: n_0' \in \mathbb{N}: f(n) \ge c'\cdot g(n) \ \forall \ n \ge n_0' $
I have to prove that: $ 0 < l < \infty$
Mi first approach was to try and use the inequalities above to reach an expression like $|\frac{f(n)}{g(n)}-l|<\epsilon$ and thus proving the exercise with the definition of a limit. But that got me nowhere, and then I realized the following: $c'\cdot g(n) \le f(n)\le c\cdot g(n) \iff c' \le \frac{f(n)}{g(n)}\le c$
Now, when $n \to \infty$, then: $c' \le l \le c \implies 0 < l < \infty $.
That's where my question enters: Is this proof correct, would I need to add something else? Is there another way to do it? Maybe using the definition of a limit?
Thanks! :)
Edit: $c,c' \in \mathbb{R_{>0}}$
Edit (2) : $g,f:\mathbb{N} \to \mathbb{R_{>0}}$
If $c'>0$, not only $c' \in \mathbb{R}$, then proof is correct: you know limit exists for $a_n=\frac{f(n)}{g(n)}$ and $c' \leqslant a_n \leqslant c$, so $c' \leqslant \lim a_n \leqslant c$.