Problems proving that $\lim\limits_{n \to \infty}\frac{2n}{n^3+1}=0 $

Question

I have to prove using the definition of a limit.

Following the definition I think I should find n for which it holds: $\lvert\frac{2n}{n^3+1}\rvert\lt\epsilon$

But after some transformations I end up with: $\frac{2}{\epsilon}\lt n^2+\frac{1}{n}$

and I don't know where to go from now on

Answer 1

HINT: Note that $${2n\over n^3 + 1} \le {2n\over n^3} = {2\over n^2}.$$

Answer 2

Note that $$ \frac{2n}{n^3+1}\leq \frac{2n}{n^3}=\frac{2}{n^2}<\epsilon $$ if $n>\sqrt{2/\epsilon}$.

Answer 3

Hint

$\dfrac{2n}{n^3+1}\leq\dfrac{2n}{n^3}=\dfrac{2}{n^2}$

Answer 4

Hint: $$\dfrac{2n}{n^3+1}=\dfrac{2}{(n+1)(n+\frac1n-1)}$$ also $$n+\frac1n\geq2$$

Answer 5

We have that

$$\left|\frac{2n}{n^3+1}\right|=\frac{2n}{n^3+1}\lt\epsilon \iff \epsilon n^3-2n+\epsilon>0$$

and for $n$ large

$$\epsilon n^3-2n+\epsilon>\epsilon n^2-2n=n(\epsilon n-2)>0 \iff n>\frac 2 \epsilon $$