# Using Limits to Determine Asymptotic Relationship

Let's say we have $$3^{4n}$$ and $$4^{3n}$$. With the note of The Asymptotic Cheat Sheet from MIT.

We should first calculate the lim n->infinity $$3^{4n}$$/$$4^{3n}$$. and it result is $$\infty$$.

In the we have

lim n→∞ f(n)/g(n) != 0,∞  ⇒  f=Θ(g)   Note 1


and

lim n→∞  f(n)/g(n)=∞     ⇒ f=ω(g)   Note 2


My questions is: Given that f(n) is $$3^{4n}$$ and g(n) is $$4^{3n}$$ in this case, which "case" should we apply ? How should one read Note 1 correctly ? Does it mean "lim n→∞ f(n)/g(n) not equal to 0, but equal to infinity? then f=Θ(g) " Or
"lim n→∞ f(n)/g(n) not equal to 0 and not equal to infinity, then f=Θ(g)?

Also if f(n) = $$\theta$$ g(n)? doesn't that already mean f(n) = $$O$$ g(n)? if so, then note 1 should read as "lim n→∞ f(n)/g(n) not equal to 0 and not equal to infinity, then f=Θ(g)?

Knowing $$3^4=81$$ and $$4^3=64$$, we have $$f(n)=81^n$$ and $$g(n)=64^n$$. Using this we obtain: $$\lim\limits_{n \to \infty}\frac{f(n)}{g(n)}=\lim\limits_{n \to \infty}\left(\frac{81}{64}\right)^n=+\infty$$ so you have your "Note 2". By definition last limit mean $$\forall C>0$$ we have $$\exists N \in \mathbb{N}$$, such that $$\forall n>N$$ holds $$\frac{f(n)}{g(n)} \geqslant C$$ so we have $$f\in \Omega(g)$$.
For further let's assume, that we have positive functions and $$g$$ have not $$0$$s subsequence.
Following question. Your "note 1" means that $$\lim\limits_{n \to \infty}\frac{f(n)}{g(n)}$$ is not $$0$$ and is not $$\infty$$, but equals to some number between them.
And last. Suppose $$\lim\limits_{n \to \infty}\frac{f(n)}{g(n)}=A>0$$. This means, that $$A-\varepsilon<\frac{f(n)}{g(n)} < A+\varepsilon$$, for appropriate $$n$$, and so we can conclude $$f \in \Theta(g)$$.