# Limit of sequence :$x_n = \frac{2n^2 + 3}{n^3 + 2n}$

Consider the sequence $x_n = \frac{2n^2 + 3}{n^3 + 2n}, n \in \mathbb{N}$. Show that $\lim_{n\to \infty} x_n = 0$

I have no idea how to find my $n_{\epsilon}$ such as $n > n_{\epsilon} \Rightarrow \left| \frac{2n^2 + 3}{n^3 + 2n} \right | < \epsilon$ . I've tried to show it is Cauchy or find another 2 sequences to use the squeeze theorem, but I had no sucess.

Can you help me to prove this (A hint would be great!)? Thanks!

Note that our expression is positive and $\lt \frac{2n^2+3n^2}{n^3}=\frac{5}{n}$.
Now finding an $n_\epsilon$ that works should be easy.
Remark: The structure of the answer was chosen to make writing out an $\epsilon$-$N$ argument straightforward. If we are allowed to use other tools, just divide top and bottom by $n^3$. The new bottom has limit $1$, the new top has limit $0$.
• Sure, the simplification was to make the $\epsilon$ stuff easy. Alternately, for showing the limit is $0$, divide top and bottom by $n^3$. – André Nicolas Oct 16 '13 at 17:21
• Alternatively, $x_n=\frac{2}{n}-\frac{1}{n^3+2n}$ – Ivan Loh Oct 16 '13 at 17:23