Little-o proof by definition

Question

I'm trying to figure out how to prove the following but to no avail.

Given the following functions :

$f(n) = n^3 -4n$

$g(n) = 5n^2 + 3n$

I have to show that $g(n) = o(f(n))$ by definition, that is not using limits or O definitions.

Notice that we say that $f(n)$ is $o(g(n))$ if for any real constant $\epsilon > 0$, there exists an integer constant $n_0 >= 1$ such that $f(n) < \epsilon g(n)$ for every integer $n>=n_0$.

I was trying to use some calculus techniques to solve it.

Here is what I have so far:

I'm looking for $n_0 >= 1$ such that $\epsilon_0 >= (5n^2+3n)/(n^3-4n)$ for every $n>= n_0$

Can someone explain me how can I continue from here? If I solve this inequality I believe It will give me the required $n_0$ for the $\epsilon$.

Answer 1

For every $\varepsilon\gt0$, choose $n_\varepsilon\geqslant4$ such that $n_\varepsilon\geqslant8/\varepsilon$. Then, for every $n\geqslant n_\varepsilon$, $$ g(n)=5n^2+3n\leqslant8(n^2-4)=8f(n)/n\leqslant(8/n_\varepsilon)f(n)\leqslant\varepsilon f(n). $$ The only nontrivial step is the assertion that $5n^2+3n\leqslant8(n^2-4)$, but this is equivalent to the inequality $3n(n-1)\geqslant32$, which indeed holds for every $n\geqslant4$.

Answer 2

Hint as $n$ tends to infinity

$$ \frac{5n^2+3n}{n^3-4n} \sim\frac{5n^2}{n^3}=\frac{5}{n} $$