2

Use the definition of convergence of a sequence to show $$\lim_{n\rightarrow \infty} \frac {2n^2}{n^3+3}= 0$$

I understand that to do this we must show $ \frac {2n^2}{n^3+3} \leq \varepsilon$, but I'm not sure how to do that.

fosho
  • 6,334
user312679
  • 39
  • If you have access to the Squeeze Theorem, the argument can be fairly short. Do you? – 2'5 9'2 Feb 11 '16 at 16:20
  • @alex.jordan preferably use only the definition if that's what the question asks for. – fosho Feb 11 '16 at 16:22
  • Actually to be pedantic you would need to specify $n$ in the limit. You probably mean $n \to \infty$, right? – mathreadler Feb 11 '16 at 16:39
  • @mathreadler It can be argued that when $a_n$ is understood to be a sequence (as opposed to a function of real $n$ or some other object), that $\lim a_n$ is unambiguously asking for the limit as $n\to\infty$. But it would come down to convention. – 2'5 9'2 Feb 11 '16 at 19:08
  • Yes you are right, I missed the word "sequence" and only saw the word limit ( at the time of reading ! ). Probably the natural numbers are the convention if the word "sequence" is used, I agree there ;) – mathreadler Feb 12 '16 at 20:55

1 Answers1

3

Given $\epsilon>0$ choose $N$ large enough such that $\frac{2}{N} < \epsilon$ so that for $n>N$ we have $$\bigg|\frac{2n^2}{n^3+3}-0\bigg|=\frac{2n^2}{n^3+3}\leq\frac{2n^2}{n^3} =\frac{2}{n}< \frac{2}{N}<\epsilon$$

fosho
  • 6,334