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 $0$.

  • $\begingroup$ Nice, but if I don't need to use the definition, can I use this facts directly applying the squeeze theorem, right? $\endgroup$ – Giiovanna Oct 16 '13 at 17:18
  • $\begingroup$ Sure, the simplification was to make the $\epsilon$ stuff easy. Alternately, for showing the limit is $0$, divide top and bottom by $n^3$. $\endgroup$ – André Nicolas Oct 16 '13 at 17:21
  • $\begingroup$ Right, like we do usually in calculus. Thanks. $\endgroup$ – Giiovanna Oct 16 '13 at 17:23
  • $\begingroup$ Alternatively, $x_n=\frac{2}{n}-\frac{1}{n^3+2n}$ $\endgroup$ – Ivan Loh Oct 16 '13 at 17:23
  • $\begingroup$ Thanks for your improvement. Someone had suggest use partial fractions before but he deleted his commentary. $\endgroup$ – Giiovanna Oct 16 '13 at 17:24

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