Proving Little-Oh notation by definition I need to prove by definition (show that for every $C$ there exists a $n_0$) that: $$5n^2+3n= o(n^3-4n)$$

By trying to simplify the expression I get to the point where I should prove that: $$ \frac{5n^2+3n}{n^3-4n} &lt; C$$.

Now, showing that $\lim\limits_{x\to\infty} \frac{5n^2+3n}{n^3-4n} = 0$ isn't a tough mission, but how do I prove it by definition? How do I point out specific $n_0$ for every $C$ that comes up? I'm asked to provide a tight answer and not let $n_0 = C^C$ so that the $n_0$ for sure will be large enough but to point out $n_0$ that is reasonably small.
Either you do what David suggested, or you do something like this:

Let $c &gt; 0$. Then

\begin{align}
5n^2 +3n &amp;\le c(n^3-4n) &amp;\iff \\
5n+3 &amp;\le cn^2-4c &amp;\iff \\
0 &amp;\le cn^2-5n-(3+4c)
\end{align}

Now solve for $n$ and let $n_0$ be the bigger root of that term.