how to prove $\lim_{x\to \infty}\frac{n}{n^3+1}=0$ using definition of limits? $$\lim_{n\to \infty}\frac{n}{n^3+1}=0$$


I know that to prove this i have to find a $N \in \Bbb N$

$\forall   \epsilon&gt;0$ s.t

if $n&gt;N $ then $|\frac{n}{n^3+1}-0|&lt; \epsilon$ i tried it some wayes but i couldnt find an answer
Using that $n\leq n^2$ and that $0\leq 1$ we get 

$$\frac{n}{n^3+1}\leq \frac{n^2}{n^3+1}\leq \frac{n^2}{n^3}=\frac{1}{n}$$
I see it this way. You can divide both numerator and denominator by n to get

lim(n-->inf) 1/(n^2+1/n)

1/n --> 0 in the denominator, so the trend would be 1/n^2, which as n approaches infinity, the limit approaches 0.