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The statement

$$\lim_{n\to ∞}\frac{\sqrt[3]{n^2}\sin n}{n+1}=0$$ is equivalent to

$$\forall \epsilon > 0 \quad \exists N_\epsilon \quad \mathrm{s.t.} \quad \forall n \geq N_\epsilon \quad \left|\frac{\sqrt[3]{n^2}\sin n}{n+1}-0 \right|\leq \epsilon.$$

I don't know how to express $N_\epsilon$ in terms of $\epsilon$ in this question.

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closed as unclear what you're asking by qbert, user99914, Namaste, Ethan Bolker, José Carlos Santos Jul 7 '18 at 20:35

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    $\begingroup$ There's no $x$ in your formula.... Do you mean $\lim_{n\to\infty}\cdots$? $\endgroup$ – Lord Shark the Unknown Jul 7 '18 at 2:11
  • $\begingroup$ @LordSharktheUnknown sorry, I made the mistake when I paste the code.... $\endgroup$ – Chloe Zhou Jul 7 '18 at 2:14
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    $\begingroup$ Can you use squeeze theorem? That would be a lot easier. $\endgroup$ – user122049 Jul 7 '18 at 2:19
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Notice that, for all $n$, because $|\sin n| \leq 1$ and $1/|n+1| \leq 1/|n|$, $$|f(n)| =\left|\frac{n^{2/3} \sin n}{n+1}\right| \leq \left|\frac{n^{2/3}}{n+1}\right| \leq \left| \frac{n^{2/3}}{n}\right| = \left| \frac 1 {n^{1/3}} \right|. $$ So $|f(n)| \leq \epsilon$ when $\left|n^{-1/3} \right|\leq \epsilon$, that is, when $$n \geq \frac 1 {\epsilon^3} =: N_\epsilon.$$

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    $\begingroup$ Would the down-voter care to explain himself/herself? I did nothing short of answering the OP's question. $\endgroup$ – giobrach Jul 7 '18 at 2:56
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Try to decompose the formula and see how the components behave individually as $n \to \infty$. How does $\sqrt[3]{n^2}$ behave? What about $\frac{1}{n+1}$? What effect do the oscillations of $\sin n$ have on the terms of the sequence (what is the limited range of $\sin n$)?

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    $\begingroup$ But in that case, am I still using the N-e language? I thought the question require me to express n in terms of e $\endgroup$ – Chloe Zhou Jul 7 '18 at 2:16
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    $\begingroup$ @ChloeZhou Yes! But before you dive into finding the $n$ for your $ \varepsilon$, try to think how the sequence behaves in terms of the components. Why do the terms get arbitrarily close to $0$? $\endgroup$ – Andrey Portnoy Jul 7 '18 at 2:19
  • $\begingroup$ I think you didn't get the question $\endgroup$ – Gaston Burrull Jul 7 '18 at 2:25
  • $\begingroup$ @GastónBurrull I'm referring you to Theorem 3.3c in Rudin's PMA. My idea is to give hints, as this is clearly a homework problem, and not just give away the solution. $\endgroup$ – Andrey Portnoy Jul 7 '18 at 2:29

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