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 $N \in \Bbb N$
$\forall   \epsilon>0$ s.t
if $n>N $ then $|\frac{n}{n^3+1}-0|< \epsilon$ i tried it some wayes but i couldnt find an answer
 A: Using that $n\leq n^2$ and that $0\leq 1$ we get 
$$\frac{n}{n^3+1}\leq \frac{n^2}{n^3+1}\leq \frac{n^2}{n^3}=\frac{1}{n}$$
A: $$\left| \frac { n }{ n^{ 3 }+1 } -0 \right| <\varepsilon \quad \left| \frac { n }{ n^{ 3 }+1 }  \right| <\left| \frac { n }{ { n }^{ 3 } }  \right| =\frac { 1 }{ { n }^{ 2 } } <\varepsilon $$ this means $$n>\frac { 1 }{ \sqrt { \varepsilon  }  } $$  where ${ n }_{ \varepsilon  }=\frac { 1 }{ \sqrt { \varepsilon  }  } $
so

$$\forall \varepsilon >0,\exists { n }_{ \varepsilon  }\epsilon N,n> { n }_{ \varepsilon  }\quad \left| \frac { n }{ n^{ 3 }+1 }  \right| <\varepsilon $$ 

A: I got an answer.
$|\frac{n}{n^3+1}|<\frac{1}{n}$ I choose $N=\lceil\frac{1}{\epsilon} \rceil$
$\forall \epsilon>0$ if $n>N \Rightarrow n>\lceil\frac{1}{\epsilon} \rceil>\frac{1}{\epsilon} \Rightarrow \epsilon>\frac{1}{n}>\frac{n}{n^3+1}>|\frac{n}{n^3+1}-0|$
A: 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.
